Chapter 5 Arithmetic Progressions

Solution:

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The given Arithmetic Progression (AP) is: $\frac{3}{2}$, $\frac{1}{2}$, $-\frac{1}{2}$, $-\frac{3}{2}$, ...

The first term of an AP is the first number in the sequence.

First term, $a = \frac{3}{2}$.

The common difference of an AP is the difference between any term and its preceding term.

Let's calculate the difference between the second term ($a_2$) and the first term ($a_1$):

$d = a_2 - a_1$

$d = \frac{1}{2} - \frac{3}{2}$

$d = \frac{1 - 3}{2}$

$d = \frac{-2}{2}$

$d = -1$

Alternatively, we can verify by calculating the difference between the third term ($a_3$) and the second term ($a_2$):

$d = a_3 - a_2$

$d = -\frac{1}{2} - \frac{1}{2}$

$d = \frac{-1 - 1}{2}$

$d = \frac{-2}{2}$

$d = -1$

The common difference is consistently $-1$.

Common difference, $d = -1$.

Thus, the first term of the given AP is $\frac{3}{2}$ and the common difference is $-1$.

A list of numbers $a_1, a_2, a_3, ...$ forms an Arithmetic Progression (AP) if the difference between consecutive terms is constant, i.e., $a_{k+1} - a_k$ is the same for all $k \ge 1$. This constant difference is called the common difference, $d$.

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(i) 4, 10, 16, 22, . . .

Let the terms be $a_1 = 4$, $a_2 = 10$, $a_3 = 16$, $a_4 = 22$.

Calculate the difference between consecutive terms:

$a_2 - a_1 = 10 - 4 = 6$

$a_3 - a_2 = 16 - 10 = 6$

$a_4 - a_3 = 22 - 16 = 6$

Since the difference between consecutive terms is constant (6), this list of numbers forms an AP.

The first term is $a = 4$ and the common difference is $d = 6$.

The next two terms are:

$a_5 = a_4 + d = 22 + 6 = 28$

$a_6 = a_5 + d = 28 + 6 = 34$

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(ii) 1, – 1, – 3, – 5, . . .

Let the terms be $a_1 = 1$, $a_2 = -1$, $a_3 = -3$, $a_4 = -5$.

Calculate the difference between consecutive terms:

$a_2 - a_1 = -1 - 1 = -2$

$a_3 - a_2 = -3 - (-1) = -3 + 1 = -2$

$a_4 - a_3 = -5 - (-3) = -5 + 3 = -2$

Since the difference between consecutive terms is constant (-2), this list of numbers forms an AP.

The first term is $a = 1$ and the common difference is $d = -2$.

The next two terms are:

$a_5 = a_4 + d = -5 + (-2) = -7$

$a_6 = a_5 + d = -7 + (-2) = -9$

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(iii) – 2, 2, – 2, 2, – 2, . . .

Let the terms be $a_1 = -2$, $a_2 = 2$, $a_3 = -2$, $a_4 = 2$, $a_5 = -2$.

Calculate the difference between consecutive terms:

$a_2 - a_1 = 2 - (-2) = 2 + 2 = 4$

$a_3 - a_2 = -2 - 2 = -4$

Since the difference between consecutive terms is not constant (4 and -4), this list of numbers does not form an AP.

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(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Let the terms be $a_1 = 1$, $a_2 = 1$, $a_3 = 1$, $a_4 = 2$, etc.

Calculate the difference between consecutive terms:

$a_2 - a_1 = 1 - 1 = 0$

$a_3 - a_2 = 1 - 1 = 0$

$a_4 - a_3 = 2 - 1 = 1$

Since the difference between consecutive terms is not constant (0 and 1), this list of numbers does not form an AP.

A list of numbers forms an Arithmetic Progression (AP) if the difference between consecutive terms is constant.

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(i) Taxi Fare

Fare for the first km = $\textsf{₹}15$ ($a_1$).

Fare for the first 2 km = Fare for 1st km + Fare for additional 1 km = $15 + 8 = \textsf{₹}23$ ($a_2$).

Fare for the first 3 km = Fare for 2 km + Fare for additional 1 km = $23 + 8 = \textsf{₹}31$ ($a_3$).

Fare for the first 4 km = Fare for 3 km + Fare for additional 1 km = $31 + 8 = \textsf{₹}39$ ($a_4$).

The list of taxi fares (total fare after each km) is 15, 23, 31, 39, ...

Let's check the difference between consecutive terms:

$a_2 - a_1 = 23 - 15 = 8$

$a_3 - a_2 = 31 - 23 = 8$

$a_4 - a_3 = 39 - 31 = 8$

Since the difference between consecutive terms is constant (8), this list of numbers makes an Arithmetic Progression.

Reason: The fare increases by a fixed amount ($\textsf{₹}8$) for each additional km.

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(ii) Amount of Air in a Cylinder

Let the initial amount of air in the cylinder be $V$ units ($a_1$).

In the first step, the vacuum pump removes $\frac{1}{4}$ of the air remaining, which is $\frac{1}{4}V$.

Amount of air remaining after the first step = $V - \frac{1}{4}V = \frac{3}{4}V$ ($a_2$).

In the second step, the pump removes $\frac{1}{4}$ of the air remaining, which is $\frac{1}{4} \left(\frac{3}{4}V\right) = \frac{3}{16}V$.

Amount of air remaining after the second step = $\frac{3}{4}V - \frac{3}{16}V = \frac{12V - 3V}{16} = \frac{9}{16}V$ ($a_3$).

In the third step, the pump removes $\frac{1}{4}$ of the air remaining, which is $\frac{1}{4} \left(\frac{9}{16}V\right) = \frac{9}{64}V$.

Amount of air remaining after the third step = $\frac{9}{16}V - \frac{9}{64}V = \frac{36V - 9V}{64} = \frac{27}{64}V$ ($a_4$).

The list of amounts of air remaining is $V, \frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, ...$

Let's check the difference between consecutive terms:

$a_2 - a_1 = \frac{3}{4}V - V = -\frac{1}{4}V$

$a_3 - a_2 = \frac{9}{16}V - \frac{3}{4}V = \frac{9V - 12V}{16} = -\frac{3}{16}V$

Since the difference between consecutive terms is not constant ($-\frac{1}{4}V \neq -\frac{3}{16}V$), this list of numbers does not make an AP.

Reason: The amount of air removed at each step is a fraction of the remaining air, not a fixed amount. The terms form a Geometric Progression ($a_k = V \left(\frac{3}{4}\right)^{k-1}$).

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(iii) Cost of Digging a Well

Cost for the first metre = $\textsf{₹}150$ ($a_1$).

Cost for the first 2 metres = Cost for 1st metre + Cost for 2nd metre = $150 + 50 = \textsf{₹}200$ ($a_2$).

Cost for the first 3 metres = Cost for 2 metres + Cost for 3rd metre = $200 + 50 = \textsf{₹}250$ ($a_3$).

Cost for the first 4 metres = Cost for 3 metres + Cost for 4th metre = $250 + 50 = \textsf{₹}300$ ($a_4$).

The list of costs (total cost after digging each metre) is 150, 200, 250, 300, ...

Let's check the difference between consecutive terms:

$a_2 - a_1 = 200 - 150 = 50$

$a_3 - a_2 = 250 - 200 = 50$

$a_4 - a_3 = 300 - 250 = 50$

Since the difference between consecutive terms is constant (50), this list of numbers makes an Arithmetic Progression.

Reason: The cost increases by a fixed amount ($\textsf{₹}50$) for each subsequent metre of digging.

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(iv) Amount of Money at Compound Interest

Principal amount deposited = $\textsf{₹}10000$. Rate of interest = 8% per annum, compounded annually.

Amount at the end of 1st year = $10000 \left(1 + \frac{8}{100}\right)^1 = 10000(1.08) = \textsf{₹}10800$ ($a_1$ is usually the amount at the start, but the question asks for the amount *every year*. Let's consider the amount at the end of each year, starting from year 1).

Let the initial deposit be $P = 10000$. Let the amount at the end of year $n$ be $A_n$.

$A_0 = 10000$ (Amount at the beginning)

Amount at the end of year 1, $A_1 = P \left(1 + \frac{r}{100}\right)^1 = 10000 \left(1 + \frac{8}{100}\right) = 10000(1.08) = 10800$.

Amount at the end of year 2, $A_2 = P \left(1 + \frac{r}{100}\right)^2 = 10000 (1.08)^2 = 10000(1.1664) = 11664$.

Amount at the end of year 3, $A_3 = P \left(1 + \frac{r}{100}\right)^3 = 10000 (1.08)^3 = 10000(1.259712) \approx 12597.12$.

The list of amounts in the account every year (starting from year 1) is 10800, 11664, 12597.12, ...

Let's check the difference between consecutive terms:

$A_2 - A_1 = 11664 - 10800 = 864$

$A_3 - A_2 = 12597.12 - 11664 = 933.12$

Since the difference between consecutive terms is not constant (864 $\neq$ 933.12), this list of numbers does not make an AP.

Reason: Compound interest adds interest to the principal amount, so the interest earned each year increases. The amounts form a Geometric Progression ($A_n = P(1+r/100)^n$).

The first four terms of an AP are given by $a_1 = a$, $a_2 = a + d$, $a_3 = a + 2d$, and $a_4 = a + 3d$.

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(i) a = 10, d = 10

$a_1 = a = 10$

$a_2 = a + d = 10 + 10 = 20$

$a_3 = a + 2d = 10 + 2(10) = 10 + 20 = 30$

$a_4 = a + 3d = 10 + 3(10) = 10 + 30 = 40$

The first four terms are 10, 20, 30, 40.

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(ii) a = –2, d = 0

$a_1 = a = -2$

$a_2 = a + d = -2 + 0 = -2$

$a_3 = a + 2d = -2 + 2(0) = -2 + 0 = -2$

$a_4 = a + 3d = -2 + 3(0) = -2 + 0 = -2$

The first four terms are -2, -2, -2, -2. (This is a constant AP)

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(iii) a = 4, d = – 3

$a_1 = a = 4$

$a_2 = a + d = 4 + (-3) = 4 - 3 = 1$

$a_3 = a + 2d = 4 + 2(-3) = 4 - 6 = -2$

$a_4 = a + 3d = 4 + 3(-3) = 4 - 9 = -5$

The first four terms are 4, 1, -2, -5.

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(iv) a = – 1, d = $\frac{1}{2}$

$a_1 = a = -1$

$a_2 = a + d = -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2}$

$a_3 = a + 2d = -1 + 2\left(\frac{1}{2}\right) = -1 + 1 = 0$

$a_4 = a + 3d = -1 + 3\left(\frac{1}{2}\right) = -1 + \frac{3}{2} = -\frac{2}{2} + \frac{3}{2} = \frac{1}{2}$

The first four terms are -1, $-\frac{1}{2}$, 0, $\frac{1}{2}$.

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(v) a = – 1.25, d = – 0.25

$a_1 = a = -1.25$

$a_2 = a + d = -1.25 + (-0.25) = -1.25 - 0.25 = -1.50$

$a_3 = a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75$

$a_4 = a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00$

The first four terms are -1.25, -1.50, -1.75, -2.00.

For an Arithmetic Progression (AP), the first term is the initial value, and the common difference is the constant difference between consecutive terms.

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(i) 3, 1, – 1, – 3, . . .

The first term is the first number in the sequence.

First term, $a = 3$.

The common difference $d$ is the difference between any term and its preceding term.

$d = \text{Second term} - \text{First term} = 1 - 3 = -2$

Check with other terms:

$d = \text{Third term} - \text{Second term} = -1 - 1 = -2$

$d = \text{Fourth term} - \text{Third term} = -3 - (-1) = -3 + 1 = -2$

The common difference is $-2$.

First term $a = 3$, Common difference $d = -2$.

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(ii) – 5, – 1, 3, 7, . . .

The first term is $-5$.

First term, $a = -5$.

The common difference $d$ is the difference between any term and its preceding term.

$d = \text{Second term} - \text{First term} = -1 - (-5) = -1 + 5 = 4$

Check with other terms:

$d = \text{Third term} - \text{Second term} = 3 - (-1) = 3 + 1 = 4$

$d = \text{Fourth term} - \text{Third term} = 7 - 3 = 4$

The common difference is $4$.

First term $a = -5$, Common difference $d = 4$.

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(iii) $\frac{1}{3}$ , $\frac{5}{3}$ , $\frac{9}{3}$ , $\frac{13}{3}$ , …

The first term is $\frac{1}{3}$.

First term, $a = \frac{1}{3}$.

The common difference $d$ is the difference between any term and its preceding term.

$d = \text{Second term} - \text{First term} = \frac{5}{3} - \frac{1}{3} = \frac{5-1}{3} = \frac{4}{3}$

Check with other terms:

$d = \text{Third term} - \text{Second term} = \frac{9}{3} - \frac{5}{3} = \frac{9-5}{3} = \frac{4}{3}$

$d = \text{Fourth term} - \text{Third term} = \frac{13}{3} - \frac{9}{3} = \frac{13-9}{3} = \frac{4}{3}$

The common difference is $\frac{4}{3}$.

First term $a = \frac{1}{3}$, Common difference $d = \frac{4}{3}$.

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(iv) 0.6, 1.7, 2.8, 3.9, . . .

The first term is $0.6$.

First term, $a = 0.6$.

The common difference $d$ is the difference between any term and its preceding term.

$d = \text{Second term} - \text{First term} = 1.7 - 0.6 = 1.1$

Check with other terms:

$d = \text{Third term} - \text{Second term} = 2.8 - 1.7 = 1.1$

$d = \text{Fourth term} - \text{Third term} = 3.9 - 2.8 = 1.1$

The common difference is $1.1$.

First term $a = 0.6$, Common difference $d = 1.1$.

To check if a list of numbers forms an AP, we find the difference between consecutive terms. If the difference is constant, it is an AP.

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(i) 2, 4, 8, 16, . . .

$a_2 - a_1 = 4 - 2 = 2$

$a_3 - a_2 = 8 - 4 = 4$

Since $a_2 - a_1 \neq a_3 - a_2$, the difference is not constant.

Therefore, this list of numbers does not form an AP.

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(ii) 2, $\frac{5}{2}$ , 3, $\frac{7}{2}$ , …

$a_2 - a_1 = \frac{5}{2} - 2 = \frac{5 - 4}{2} = \frac{1}{2}$

$a_3 - a_2 = 3 - \frac{5}{2} = \frac{6 - 5}{2} = \frac{1}{2}$

$a_4 - a_3 = \frac{7}{2} - 3 = \frac{7 - 6}{2} = \frac{1}{2}$

Since the difference between consecutive terms is constant ($\frac{1}{2}$), this list of numbers forms an AP.

Common difference, $d = \frac{1}{2}$.

The next three terms are:

$a_5 = a_4 + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$

$a_6 = a_5 + d = 4 + \frac{1}{2} = \frac{8 + 1}{2} = \frac{9}{2}$

$a_7 = a_6 + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$

The next three terms are 4, $\frac{9}{2}$, 5.

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(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

$a_2 - a_1 = -3.2 - (-1.2) = -3.2 + 1.2 = -2.0$

$a_3 - a_2 = -5.2 - (-3.2) = -5.2 + 3.2 = -2.0$

$a_4 - a_3 = -7.2 - (-5.2) = -7.2 + 5.2 = -2.0$

Since the difference between consecutive terms is constant ($-2.0$), this list of numbers forms an AP.

Common difference, $d = -2.0$.

The next three terms are:

$a_5 = a_4 + d = -7.2 + (-2.0) = -9.2$

$a_6 = a_5 + d = -9.2 + (-2.0) = -11.2$

$a_7 = a_6 + d = -11.2 + (-2.0) = -13.2$

The next three terms are -9.2, -11.2, -13.2.

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(iv) – 10, – 6, – 2, 2, . . .

$a_2 - a_1 = -6 - (-10) = -6 + 10 = 4$

$a_3 - a_2 = -2 - (-6) = -2 + 6 = 4$

$a_4 - a_3 = 2 - (-2) = 2 + 2 = 4$

Since the difference between consecutive terms is constant (4), this list of numbers forms an AP.

Common difference, $d = 4$.

The next three terms are:

$a_5 = a_4 + d = 2 + 4 = 6$

$a_6 = a_5 + d = 6 + 4 = 10$

$a_7 = a_6 + d = 10 + 4 = 14$

The next three terms are 6, 10, 14.

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(v) 3, $3 + \sqrt{2}$ , $3 + 2\sqrt{2}$ , $3 + 3\sqrt{2}$ , . . .

$a_2 - a_1 = (3 + \sqrt{2}) - 3 = \sqrt{2}$

$a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = 3 + 2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2}$

$a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = 3 + 3\sqrt{2} - 3 - 2\sqrt{2} = \sqrt{2}$

Since the difference between consecutive terms is constant ($\sqrt{2}$), this list of numbers forms an AP.

Common difference, $d = \sqrt{2}$.

The next three terms are:

$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$

$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$

$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$

The next three terms are $3 + 4\sqrt{2}$, $3 + 5\sqrt{2}$, $3 + 6\sqrt{2}$.

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(vi) 0.2, 0.22, 0.222, 0.2222, . . .

$a_2 - a_1 = 0.22 - 0.2 = 0.02$

$a_3 - a_2 = 0.222 - 0.22 = 0.002$

Since $a_2 - a_1 \neq a_3 - a_2$, the difference is not constant.

Therefore, this list of numbers does not form an AP.

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(vii) 0, – 4, – 8, –12, . . .

$a_2 - a_1 = -4 - 0 = -4$

$a_3 - a_2 = -8 - (-4) = -8 + 4 = -4$

$a_4 - a_3 = -12 - (-8) = -12 + 8 = -4$

Since the difference between consecutive terms is constant ($-4$), this list of numbers forms an AP.

Common difference, $d = -4$.

The next three terms are:

$a_5 = a_4 + d = -12 + (-4) = -16$

$a_6 = a_5 + d = -16 + (-4) = -20$

$a_7 = a_6 + d = -20 + (-4) = -24$

The next three terms are -16, -20, -24.

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(viii) $-\frac{1}{2}$ , $-\frac{1}{2}$ , $-\frac{1}{2}$ , $-\frac{1}{2}$ …

$a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

$a_3 - a_2 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

$a_4 - a_3 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Since the difference between consecutive terms is constant (0), this list of numbers forms an AP.

Common difference, $d = 0$.

The next three terms are:

$a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

$a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

$a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

The next three terms are $-\frac{1}{2}$, $-\frac{1}{2}$, $-\frac{1}{2}$.

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(ix) 1, 3, 9, 27, . . .

$a_2 - a_1 = 3 - 1 = 2$

$a_3 - a_2 = 9 - 3 = 6$

Since $a_2 - a_1 \neq a_3 - a_2$, the difference is not constant.

Therefore, this list of numbers does not form an AP.

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(x) a, 2a, 3a, 4a, . . .

$a_2 - a_1 = 2a - a = a$

$a_3 - a_2 = 3a - 2a = a$

$a_4 - a_3 = 4a - 3a = a$

The difference between consecutive terms is constant ($a$). This list of numbers forms an AP (assuming $a$ is a fixed value).

Common difference, $d = a$.

The next three terms are:

$a_5 = a_4 + d = 4a + a = 5a$

$a_6 = a_5 + d = 5a + a = 6a$

$a_7 = a_6 + d = 6a + a = 7a$

The next three terms are 5a, 6a, 7a.

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(xi) a, $a^2$ , $a^3$ , $a^4$ , . . .

$a_2 - a_1 = a^2 - a = a(a - 1)$

$a_3 - a_2 = a^3 - a^2 = a^2(a - 1)$

The difference $a(a-1)$ is equal to $a^2(a-1)$ only if $a=0$ or $a=1$. For other values of $a$, the difference is not constant.

In general, this list of numbers does not form an AP.

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(xii) $\sqrt{2}$ , $\sqrt{8}$ , $\sqrt{18}$ , $\sqrt{32}$ , . . .

Rewrite the terms in simplest radical form:

$\sqrt{2}$

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

The sequence is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, ...$

$a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$

$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

$a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$

Since the difference between consecutive terms is constant ($\sqrt{2}$), this list of numbers forms an AP.

Common difference, $d = \sqrt{2}$.

The next three terms are:

$a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$

$a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$

$a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$

The next three terms are $\sqrt{50}$, $\sqrt{72}$, $\sqrt{98}$.

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(xiii) $\sqrt{3}$ , $\sqrt{6}$ , $\sqrt{9}$ , $\sqrt{12}$ , . . .

Rewrite the terms in simplest radical form:

$\sqrt{3}$

$\sqrt{6}$

$\sqrt{9} = 3$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

The sequence is $\sqrt{3}, \sqrt{6}, 3, 2\sqrt{3}, ...$

$a_2 - a_1 = \sqrt{6} - \sqrt{3}$

$a_3 - a_2 = 3 - \sqrt{6}$

Since $\sqrt{6} - \sqrt{3} \neq 3 - \sqrt{6}$ (approx values: $2.45 - 1.73 = 0.72$ and $3 - 2.45 = 0.55$), the difference is not constant.

Therefore, this list of numbers does not form an AP.

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(xiv) 1² , 3² , 5² , 7² , . . .

Evaluate the terms:

$1^2 = 1$

$3^2 = 9$

$5^2 = 25$

$7^2 = 49$

The sequence is 1, 9, 25, 49, ...

$a_2 - a_1 = 9 - 1 = 8$

$a_3 - a_2 = 25 - 9 = 16$

Since $a_2 - a_1 \neq a_3 - a_2$, the difference is not constant.

Therefore, this list of numbers does not form an AP.

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(xv) 1² , 5² , 7² , 73, . . .

Evaluate the given terms:

$a_1 = 1^2 = 1$

$a_2 = 5^2 = 25$

$a_3 = 7^2 = 49$

$a_4 = 73$

The sequence is 1, 25, 49, 73, ...

Difference between consecutive terms:

$a_2 - a_1 = 25 - 1 = 24$

$a_3 - a_2 = 49 - 25 = 24$

$a_4 - a_3 = 73 - 49 = 24$

Since the difference between consecutive terms is constant (24), this list of numbers forms an AP.

The common difference is $d = 24$.

The next three terms are:

$a_5 = a_4 + d = 73 + 24 = 97$

$a_6 = a_5 + d = 97 + 24 = 121$

$a_7 = a_6 + d = 121 + 24 = 145$

The next three terms are 97, 121, and 145.

Solution:

The given Arithmetic Progression (AP) is: 2, 7, 12, ...

From the AP, we can identify the first term ($a$) and the common difference ($d$).

First term, $a = 2$.

Common difference, $d = \text{Second term} - \text{First term} = 7 - 2 = 5$.

We need to find the 10th term of this AP. The formula for the $n$-th term of an AP is:

$a_n = a + (n - 1)d$

Here, we need to find the 10th term, so $n = 10$.

Substitute the values of $a$, $n$, and $d$ into the formula:

$a_{10} = a + (10 - 1)d$

$a_{10} = 2 + (9) \times 5$

$a_{10} = 2 + 45$

$a_{10} = 47$

Thus, the 10th term of the AP is 47.

Solution:

The given Arithmetic Progression (AP) is: 21, 18, 15, ...

The first term is $a = 21$.

The common difference is $d = a_2 - a_1 = 18 - 21 = -3$.

We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

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Finding which term is -81:

We want to find $n$ such that $a_n = -81$.

Set $a_n = -81$ in the formula:

$-81 = 21 + (n - 1)(-3)$

Subtract 21 from both sides:

$-81 - 21 = (n - 1)(-3)$

$-102 = -3(n - 1)$

Divide both sides by -3:

$\frac{-102}{-3} = n - 1$

$34 = n - 1$

Add 1 to both sides:

$n = 34 + 1$

$n = 35$

Since $n = 35$ is a positive integer, -81 is the 35th term of the AP.

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Checking if any term is 0:

We want to find if there is a positive integer $n$ such that $a_n = 0$.

Set $a_n = 0$ in the formula:

$0 = 21 + (n - 1)(-3)$

Subtract 21 from both sides:

$0 - 21 = (n - 1)(-3)$

$-21 = -3(n - 1)$

Divide both sides by -3:

$\frac{-21}{-3} = n - 1$

$7 = n - 1$

Add 1 to both sides:

$n = 7 + 1$

$n = 8$

Since $n = 8$ is a positive integer, 0 is the 8th term of the AP.

---

Reason:

We found that setting $a_n = -81$ gives $n=35$, which is a positive integer. This indicates that -81 is a term in the sequence.

Similarly, setting $a_n = 0$ gives $n=8$, which is also a positive integer. This indicates that 0 is a term in the sequence.

Specifically, the 35th term is -81, and the 8th term is 0.

Solution:

Let the first term of the AP be $a$ and the common difference be $d$.

The formula for the $n$-th term of an AP is $a_n = a + (n - 1)d$.

We are given that the 3rd term is 5. Using the formula with $n=3$:

$a_3 = a + (3 - 1)d$

We are also given that the 7th term is 9. Using the formula with $n=7$:

$a_7 = a + (7 - 1)d$

Now we have a system of two linear equations with variables $a$ and $d$:

$a + 2d = 5$

$a + 6d = 9$

Subtract equation (1) from equation (2) to eliminate $a$:

$(a + 6d) - (a + 2d) = 9 - 5$

$a + 6d - a - 2d = 4$

$(a - a) + (6d - 2d) = 4$

$0 + 4d = 4$

$4d = 4$

$d = \frac{4}{4}$

$d = 1$

Now substitute the value of $d = 1$ into equation (1) to find $a$:

$a + 2(1) = 5$

$a + 2 = 5$

$a = 5 - 2$

$a = 3$

The first term is $a = 3$ and the common difference is $d = 1$.

The AP is formed by starting with the first term and adding the common difference repeatedly.

First term: $a_1 = a = 3$

Second term: $a_2 = a + d = 3 + 1 = 4$

Third term: $a_3 = a + 2d = 3 + 2(1) = 3 + 2 = 5$ (Matches the given information)

Fourth term: $a_4 = a + 3d = 3 + 3(1) = 3 + 3 = 6$

Fifth term: $a_5 = a + 4d = 3 + 4(1) = 3 + 4 = 7$

Sixth term: $a_6 = a + 5d = 3 + 5(1) = 3 + 5 = 8$

Seventh term: $a_7 = a + 6d = 3 + 6(1) = 3 + 6 = 9$ (Matches the given information)

The AP is 3, 4, 5, 6, 7, 8, 9, ...

Thus, the AP is $3, 4, 5, 6, \dots$

Solution:

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The given list of numbers is 5, 11, 17, 23, ...

Let's first check if this list forms an Arithmetic Progression (AP) by finding the difference between consecutive terms:

$11 - 5 = 6$

$17 - 11 = 6$

$23 - 17 = 6$

Since the difference between consecutive terms is constant (6), the given list of numbers forms an AP.

The first term is $a = 5$.

The common difference is $d = 6$.

We want to check if 301 is a term of this AP. Let's assume that 301 is the $n$-th term of the AP, i.e., $a_n = 301$.

We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute the values of $a$, $d$, and $a_n$ into the formula:

$301 = 5 + (n - 1)6$

Now, we solve this equation for $n$.

$301 - 5 = (n - 1)6$

$296 = 6(n - 1)$

Divide both sides by 6:

$\frac{296}{6} = n - 1$

Simplify the fraction:

$\frac{\cancel{296}^{148}}{\cancel{6}_3} = n - 1$

$\frac{148}{3} = n - 1$

Add 1 to both sides:

$n = \frac{148}{3} + 1$

$n = \frac{148}{3} + \frac{3}{3}$

$n = \frac{148 + 3}{3}$

$n = \frac{151}{3}$

For 301 to be a term of the AP, the value of $n$ must be a positive integer (1, 2, 3, ...). Since $n = \frac{151}{3}$ is not an integer, 301 is not a term of the given AP.

Thus, 301 is not a term of the list of numbers 5, 11, 17, 23, ...

Solution:

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The two-digit numbers are the integers from 10 to 99.

We are looking for two-digit numbers that are divisible by 3.

The first two-digit number divisible by 3 is the smallest multiple of 3 that is 10 or greater. $3 \times 3 = 9$, which is not a two-digit number. $3 \times 4 = 12$, which is a two-digit number. So, the first term of our sequence is 12.

The subsequent two-digit numbers divisible by 3 will form an Arithmetic Progression with a common difference of 3.

The last two-digit number divisible by 3 is the largest multiple of 3 that is 99 or less. $99 \div 3 = 33$. So, 99 is divisible by 3. The last term of our sequence is 99.

The list of two-digit numbers divisible by 3 is: 12, 15, 18, ..., 99.

This is an AP with:

First term, $a = 12$.

Common difference, $d = 3$.

Last term, $a_n = 99$.

We need to find the number of terms ($n$) in this AP. We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute the known values into the formula:

$99 = 12 + (n - 1)3$

Subtract 12 from both sides:

$99 - 12 = (n - 1)3$

$87 = 3(n - 1)$

Divide both sides by 3:

$\frac{87}{3} = n - 1$

$29 = n - 1$

Add 1 to both sides:

$n = 29 + 1$

$n = 30$

Since $n = 30$ is a positive integer, it is the number of terms in the AP.

Thus, there are 30 two-digit numbers divisible by 3.

Solution:

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The given Arithmetic Progression (AP) is: 10, 7, 4, ..., – 62.

The first term is $a = 10$.

The common difference is $d = a_2 - a_1 = 7 - 10 = -3$.

The last term of the AP is $l = -62$.

We want to find the 11th term from the last term.

One way to do this is to consider the AP in reverse order. When the AP is reversed, the last term becomes the first term, and the common difference becomes the negative of the original common difference.

The reversed AP starts with $-62$.

The common difference of the reversed AP is $-d = -(-3) = 3$.

The reversed AP is: – 62, – 59, – 56, ..., 10.

The 11th term from the last of the original AP is the 11th term from the beginning of the reversed AP.

For the reversed AP, the first term is $a' = -62$ and the common difference is $d' = 3$.

We need to find the 11th term ($a'_{11}$) of this reversed AP using the formula $a'_n = a' + (n - 1)d'$.

Substitute $n = 11$, $a' = -62$, and $d' = 3$:

$a'_{11} = -62 + (11 - 1) \times 3$

$a'_{11} = -62 + (10) \times 3$

$a'_{11} = -62 + 30$

$a'_{11} = -32$

Thus, the 11th term from the last term of the given AP is -32.

Solution:

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Principal amount $P = \textsf{₹}1000$.

Rate of simple interest $R = 8\%$ per year.

The formula for simple interest (SI) is $SI = \frac{P \times R \times T}{100}$, where $T$ is the time in years.

Calculate the interest at the end of each year:

Interest at the end of 1st year ($T=1$): $SI_1 = \frac{1000 \times 8 \times 1}{100} = \frac{8000}{100} = \textsf{₹}80$.

Interest at the end of 2nd year ($T=2$): $SI_2 = \frac{1000 \times 8 \times 2}{100} = \frac{16000}{100} = \textsf{₹}160$.

Interest at the end of 3rd year ($T=3$): $SI_3 = \frac{1000 \times 8 \times 3}{100} = \frac{24000}{100} = \textsf{₹}240$.

Interest at the end of 4th year ($T=4$): $SI_4 = \frac{1000 \times 8 \times 4}{100} = \frac{32000}{100} = \textsf{₹}320$.

The list of interests at the end of each year is 80, 160, 240, 320, ...

Check if this list forms an AP by finding the difference between consecutive terms:

Difference between 2nd and 1st year's interest = $160 - 80 = 80$.

Difference between 3rd and 2nd year's interest = $240 - 160 = 80$.

Difference between 4th and 3rd year's interest = $320 - 240 = 80$.

Since the difference between consecutive terms is constant (80), the interests at the end of each year form an AP.

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Finding the interest at the end of 30 years:

The sequence of interests is an AP with:

First term, $a = 80$ (Interest at the end of 1st year)

Common difference, $d = 80$ (The simple interest for each additional year is constant).

We need to find the interest at the end of 30 years. This corresponds to the 30th term of the AP ($a_{30}$).

Using the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute $n = 30$, $a = 80$, and $d = 80$:

$a_{30} = 80 + (30 - 1) \times 80$

$a_{30} = 80 + (29) \times 80$

$a_{30} = 80 + 2320$

$a_{30} = 2400$

Thus, the interest at the end of 30 years is $\textsf{₹}2400$.

Alternatively, using the simple interest formula directly for $T=30$ years:

$SI_{30} = \frac{1000 \times 8 \times 30}{100} = \frac{240000}{100} = \textsf{₹}2400$. This confirms the result obtained using the AP concept.

Solution:

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The number of rose plants in the rows are 23, 21, 19, ..., 5.

Let's check if this list of numbers forms an Arithmetic Progression (AP) by finding the difference between consecutive terms:

Difference between the number of plants in the 2nd and 1st row = $21 - 23 = -2$.

Difference between the number of plants in the 3rd and 2nd row = $19 - 21 = -2$.

Since the difference is constant (-2), this list of numbers forms an AP.

The first term is the number of plants in the first row, $a = 23$.

The common difference is $d = -2$.

The last row has 5 rose plants, so the last term of the AP is $a_n = 5$.

We need to find the number of rows, which is the number of terms ($n$) in this AP.

We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute the values of $a$, $d$, and $a_n$ into the formula:

$5 = 23 + (n - 1)(-2)$

Subtract 23 from both sides:

$5 - 23 = (n - 1)(-2)$

$-18 = -2(n - 1)$

Divide both sides by -2:

$\frac{-18}{-2} = n - 1$

$9 = n - 1$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

Since $n = 10$ is a positive integer, it represents the number of rows in the flower bed.

Thus, there are 10 rows in the flower bed.

p>The formula for the $n$-th term of an Arithmetic Progression (AP) is given by $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the term number.

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(i) Given: $a = 7$, $d = 3$, $n = 8$. Find $a_n$.

Using the formula $a_n = a + (n-1)d$:

$a_8 = 7 + (8-1) \times 3$

$a_8 = 7 + (7) \times 3$

$a_8 = 7 + 21$

$a_8 = 28$

Thus, $a_n = 28$.

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(ii) Given: $a = -18$, $n = 10$, $a_n = 0$. Find $d$.

Using the formula $a_n = a + (n-1)d$:

$0 = -18 + (10-1)d$

$0 = -18 + 9d$

Add 18 to both sides:

$18 = 9d$

Divide by 9:

$d = \frac{18}{9}$

$d = 2$

Thus, $d = 2$.

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(iii) Given: $d = -3$, $n = 18$, $a_n = -5$. Find $a$.

Using the formula $a_n = a + (n-1)d$:

$-5 = a + (18-1) \times (-3)$

$-5 = a + (17) \times (-3)$

$-5 = a - 51$

Add 51 to both sides:

$a = -5 + 51$

$a = 46$

Thus, $a = 46$.

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(iv) Given: $a = -18.9$, $d = 2.5$, $a_n = 3.6$. Find $n$.

Using the formula $a_n = a + (n-1)d$:

$3.6 = -18.9 + (n-1) \times 2.5$

Add 18.9 to both sides:

$3.6 + 18.9 = (n-1) \times 2.5$

$22.5 = 2.5(n-1)$

Divide by 2.5:

$\frac{22.5}{2.5} = n-1$

$9 = n-1$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

Thus, $n = 10$.

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(v) Given: $a = 3.5$, $d = 0$, $n = 105$. Find $a_n$.

Using the formula $a_n = a + (n-1)d$:

$a_{105} = 3.5 + (105-1) \times 0$

$a_{105} = 3.5 + (104) \times 0$

$a_{105} = 3.5 + 0$

$a_{105} = 3.5$

Thus, $a_n = 3.5$.

We use the formula for the $n$-th term of an AP: $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

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(i) 30th term of the AP: 10, 7, 4, . . .

The first term is $a = 10$.

The common difference is $d = \text{Second term} - \text{First term} = 7 - 10 = -3$.

We need to find the 30th term, so $n = 30$.

Using the formula $a_n = a + (n-1)d$:

$a_{30} = 10 + (30-1)(-3)$

$a_{30} = 10 + (29)(-3)$

$a_{30} = 10 - 87$

$a_{30} = -77$

The 30th term of the AP is -77.

The correct choice is (C) –77.

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(ii) 11th term of the AP: – 3 , $-\frac{1}{2}$ , 2 ,…..

The first term is $a = -3$.

The common difference is $d = \text{Second term} - \text{First term} = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = -\frac{1}{2} + \frac{6}{2} = \frac{-1 + 6}{2} = \frac{5}{2}$.

Check with the third term: $2 - (-\frac{1}{2}) = 2 + \frac{1}{2} = \frac{4+1}{2} = \frac{5}{2}$. The common difference is indeed $\frac{5}{2}$.

We need to find the 11th term, so $n = 11$.

Using the formula $a_n = a + (n-1)d$:

$a_{11} = -3 + (11-1)\left(\frac{5}{2}\right)$

$a_{11} = -3 + (10)\left(\frac{5}{2}\right)$

$a_{11} = -3 + \cancel{10}^5 \times \frac{5}{\cancel{2}_1}$

$a_{11} = -3 + 5 \times 5$

$a_{11} = -3 + 25$

$a_{11} = 22$

The 11th term of the AP is 22.

The correct choice is (B) 22.

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

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(i) 2 , $\square$ , 26

This is an AP with 3 terms. The first term is $a_1 = 2$. The third term is $a_3 = 26$. We need to find the second term, $a_2$.

Using the formula for the $n$-th term:

$a_3 = a_1 + (3-1)d$

$26 = 2 + 2d$

Subtract 2 from both sides:

$26 - 2 = 2d$

$24 = 2d$

Divide by 2:

$d = \frac{24}{2} = 12$

The common difference is 12.

The missing term is the second term, $a_2 = a_1 + d$:

$a_2 = 2 + 12 = 14$

The missing term is 14.

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(ii) $\square$ , 13 , $\square$ , 3

This is an AP with 4 terms. The second term is $a_2 = 13$. The fourth term is $a_4 = 3$. We need to find the first term $a_1$ and the third term $a_3$.

Using the formula for the $n$-th term:

$a_2 = a_1 + (2-1)d \implies 13 = a_1 + d$ ... (1)

$a_4 = a_1 + (4-1)d \implies 3 = a_1 + 3d$ ... (2)

Subtract equation (1) from equation (2):

$(a_1 + 3d) - (a_1 + d) = 3 - 13$

$2d = -10$

$d = \frac{-10}{2} = -5$

The common difference is -5.

Substitute $d = -5$ into equation (1) to find $a_1$:

$13 = a_1 + (-5)$

$13 = a_1 - 5$

$a_1 = 13 + 5 = 18$

The first term is 18.

The missing third term is $a_3 = a_1 + 2d$:

$a_3 = 18 + 2(-5) = 18 - 10 = 8$

The missing terms are 18 and 8.

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(iii) 5 , $\square$ , $\square$ , $9\frac{1}{2}$

This is an AP with 4 terms. The first term is $a_1 = 5$. The fourth term is $a_4 = 9\frac{1}{2} = \frac{18+1}{2} = \frac{19}{2}$. We need to find the second term $a_2$ and the third term $a_3$.

Using the formula for the $n$-th term:

$a_4 = a_1 + (4-1)d$

$\frac{19}{2} = 5 + 3d$

Subtract 5 from both sides:

$\frac{19}{2} - 5 = 3d$

$\frac{19}{2} - \frac{10}{2} = 3d$

$\frac{9}{2} = 3d$

Divide by 3:

$d = \frac{9/2}{3} = \frac{9}{2 \times 3} = \frac{9}{6} = \frac{3}{2}$

The common difference is $\frac{3}{2}$.

The missing terms are:

$a_2 = a_1 + d = 5 + \frac{3}{2} = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}$

$a_3 = a_1 + 2d = 5 + 2\left(\frac{3}{2}\right) = 5 + 3 = 8$

The missing terms are $\frac{13}{2}$ and 8.

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(iv) – 4 , $\square$ , $\square$ , $\square$ , $\square$ , 6

This is an AP with 6 terms. The first term is $a_1 = -4$. The sixth term is $a_6 = 6$. We need to find the terms $a_2, a_3, a_4, a_5$.

Using the formula for the $n$-th term:

$a_6 = a_1 + (6-1)d$

$6 = -4 + 5d$

Add 4 to both sides:

$6 + 4 = 5d$

$10 = 5d$

Divide by 5:

$d = \frac{10}{5} = 2$

The common difference is 2.

The missing terms are:

$a_2 = a_1 + d = -4 + 2 = -2$

$a_3 = a_1 + 2d = -4 + 2(2) = -4 + 4 = 0$

$a_4 = a_1 + 3d = -4 + 3(2) = -4 + 6 = 2$

$a_5 = a_1 + 4d = -4 + 4(2) = -4 + 8 = 4$

The missing terms are -2, 0, 2, 4.

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(v) $\square$ , 38 , $\square$ , $\square$ , $\square$ , – 22

This is an AP with 6 terms. The second term is $a_2 = 38$. The sixth term is $a_6 = -22$. We need to find the terms $a_1, a_3, a_4, a_5$.

Using the formula for the $n$-th term:

$a_2 = a_1 + (2-1)d \implies 38 = a_1 + d$ ... (1)

$a_6 = a_1 + (6-1)d \implies -22 = a_1 + 5d$ ... (2)

Subtract equation (1) from equation (2):

$(a_1 + 5d) - (a_1 + d) = -22 - 38$

$4d = -60$

$d = \frac{-60}{4} = -15$

The common difference is -15.

Substitute $d = -15$ into equation (1) to find $a_1$:

$38 = a_1 + (-15)$

$38 = a_1 - 15$

$a_1 = 38 + 15 = 53$

The first term is 53.

The missing terms are:

$a_3 = a_1 + 2d = 53 + 2(-15) = 53 - 30 = 23$

$a_4 = a_1 + 3d = 53 + 3(-15) = 53 - 45 = 8$

$a_5 = a_1 + 4d = 53 + 4(-15) = 53 - 60 = -7$

The missing terms are 53, 23, 8, -7.

Solution:

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The given Arithmetic Progression (AP) is: 3, 8, 13, 18, ...

From the AP, we identify the first term ($a$) and the common difference ($d$).

First term, $a = 3$.

Common difference, $d = \text{Second term} - \text{First term} = 8 - 3 = 5$.

We want to find which term of this AP is 78. Let the $n$-th term be 78, i.e., $a_n = 78$.

We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute the values of $a$, $d$, and $a_n$ into the formula:

$78 = 3 + (n - 1)5$

Subtract 3 from both sides:

$78 - 3 = (n - 1)5$

$75 = 5(n - 1)$

Divide both sides by 5:

$\frac{75}{5} = n - 1$

$15 = n - 1$

Add 1 to both sides:

$n = 15 + 1$

$n = 16$

Since $n = 16$ is a positive integer, 78 is the 16th term of the AP.

Thus, the 16th term of the AP is 78.

We use the formula for the $n$-th term of an AP: $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.

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(i) 7, 13, 19, . . . , 205

The first term is $a = 7$.

The common difference is $d = 13 - 7 = 6$.

The last term is $a_n = 205$.

Substitute these values into the formula $a_n = a + (n-1)d$:

$205 = 7 + (n-1)6$

Subtract 7 from both sides:

$205 - 7 = (n-1)6$

$198 = 6(n-1)$

Divide both sides by 6:

$\frac{198}{6} = n-1$

$33 = n-1$

Add 1 to both sides:

$n = 33 + 1$

$n = 34$

Thus, there are 34 terms in the AP 7, 13, 19, . . . , 205.

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(ii) 18, $15\frac{1}{2}$ , 13, . . . , – 47

The first term is $a = 18$.

The common difference is $d = 15\frac{1}{2} - 18 = \frac{31}{2} - \frac{36}{2} = \frac{31 - 36}{2} = -\frac{5}{2}$.

The last term is $a_n = -47$.

Substitute these values into the formula $a_n = a + (n-1)d$:

$-47 = 18 + (n-1)\left(-\frac{5}{2}\right)$

Subtract 18 from both sides:

$-47 - 18 = (n-1)\left(-\frac{5}{2}\right)$

$-65 = -\frac{5}{2}(n-1)$

Multiply both sides by $-\frac{2}{5}$ to isolate $(n-1)$:

$-65 \times \left(-\frac{2}{5}\right) = n-1$

$\frac{\cancel{-65}^{13} \times (-2)}{\cancel{5}_1} = n-1$

$13 \times 2 = n-1$

$26 = n-1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

Thus, there are 27 terms in the AP 18, $15\frac{1}{2}$ , 13, . . . , – 47.

Solution:

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The given Arithmetic Progression (AP) is: 11, 8, 5, 2, ...

From the AP, we identify the first term ($a$) and the common difference ($d$).

First term, $a = 11$.

Common difference, $d = \text{Second term} - \text{First term} = 8 - 11 = -3$.

We want to check if –150 is a term of this AP. Let's assume that –150 is the $n$-th term of the AP, i.e., $a_n = -150$.

We use the formula for the $n$-th term of an AP: $a_n = a + (n - 1)d$.

Substitute the values of $a$, $d$, and $a_n$ into the formula:

$-150 = 11 + (n - 1)(-3)$

Subtract 11 from both sides:

$-150 - 11 = (n - 1)(-3)$

$-161 = -3(n - 1)$

Divide both sides by -3:

$\frac{-161}{-3} = n - 1$

$\frac{161}{3} = n - 1$

Add 1 to both sides:

$n = \frac{161}{3} + 1$

$n = \frac{161}{3} + \frac{3}{3}$

$n = \frac{161 + 3}{3}$

$n = \frac{164}{3}$

For –150 to be a term of the AP, the value of $n$ must be a positive integer (1, 2, 3, ...). Since $n = \frac{164}{3}$ is not an integer, –150 is not a term of the given AP.

Thus, –150 is not a term of the AP : 11, 8, 5, 2, ...

Solution:

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Let the first term of the AP be $a$ and the common difference be $d$.

The formula for the $n$-th term of an AP is $a_n = a + (n - 1)d$.

We are given that the 11th term is 38. Using the formula with $n=11$:

$a_{11} = a + (11 - 1)d$

We are also given that the 16th term is 73. Using the formula with $n=16$:

$a_{16} = a + (16 - 1)d$

Now we have a system of two linear equations with variables $a$ and $d$:

$a + 10d = 38$

$a + 15d = 73$

Subtract equation (1) from equation (2) to eliminate $a$:

$(a + 15d) - (a + 10d) = 73 - 38$

$a + 15d - a - 10d = 35$

$(a - a) + (15d - 10d) = 35$

$0 + 5d = 35$

$5d = 35$

Divide by 5:

$d = \frac{35}{5}$

$d = 7$

Now substitute the value of $d = 7$ into equation (1) to find $a$:

$a + 10(7) = 38$

$a + 70 = 38$

$a = 38 - 70$

$a = -32$

The first term is $a = -32$ and the common difference is $d = 7$.

We need to find the 31st term of this AP, i.e., $a_{31}$.

Using the formula $a_n = a + (n-1)d$ with $n=31$:

$a_{31} = -32 + (31-1) \times 7$

$a_{31} = -32 + (30) \times 7$

$a_{31} = -32 + 210$

$a_{31} = 178$

Thus, the 31st term of the AP is 178.

Given:

An AP with 50 terms.
3rd term ($a_3$) = 12.
Last term ($a_{50}$) = 106.

To Find:

The 29th term ($a_{29}$) of the AP.

---

Solution:

Let the first term of the AP be $a$ and the common difference be $d$.

The formula for the $n$-th term of an AP is $a_n = a + (n - 1)d$.

Using the given information, we can form a pair of linear equations.

The 3rd term is 12:

$a_3 = a + (3 - 1)d$

The last term is the 50th term ($n=50$) and is 106:

$a_{50} = a + (50 - 1)d$

Now we have a system of two linear equations:

$a + 2d = 12$

$a + 49d = 106$

Subtract equation (1) from equation (2) to eliminate $a$:

$(a + 49d) - (a + 2d) = 106 - 12$

$a + 49d - a - 2d = 94$

$(a - a) + (49d - 2d) = 94$

$0 + 47d = 94$

$47d = 94$

Divide by 47:

$d = \frac{94}{47}$

$d = 2$

The common difference is 2.

Now substitute the value of $d = 2$ into equation (1) to find the first term $a$:

$12 = a + 2(2)$

$12 = a + 4$

Subtract 4 from both sides:

$a = 12 - 4$

$a = 8$

The first term is 8.

Now that we have the first term ($a=8$) and the common difference ($d=2$), we can find the 29th term ($a_{29}$) using the formula $a_n = a + (n-1)d$ with $n=29$:

$a_{29} = a + (29 - 1)d$

$a_{29} = 8 + (28) \times 2$

$a_{29} = 8 + 56$

$a_{29} = 64$

Thus, the 29th term of the AP is 64.

Given:

The 3rd term of an AP ($a_3$) is 4.

The 9th term of the same AP ($a_9$) is -8.

To Find:

Which term ($n$) of this AP is zero ($a_n = 0$).

---

Solution:

Let the first term of the AP be $a$ and the common difference be $d$.

The formula for the $n$-th term of an AP is $a_n = a + (n - 1)d$.

Using the given information, we can form a pair of linear equations.

The 3rd term is 4 ($n=3$):

$a_3 = a + (3 - 1)d$

The 9th term is -8 ($n=9$):

$a_9 = a + (9 - 1)d$

Now we have a system of two linear equations:

$a + 2d = 4$

$a + 8d = -8$

Subtract equation (1) from equation (2) to eliminate $a$:

$(a + 8d) - (a + 2d) = -8 - 4$

$a + 8d - a - 2d = -12$

$(a - a) + (8d - 2d) = -12$

$0 + 6d = -12$

$6d = -12$

Divide by 6:

$d = \frac{-12}{6}$

$d = -2$

The common difference is -2.

Now substitute the value of $d = -2$ into equation (1) to find the first term $a$:

$4 = a + 2(-2)$

$4 = a - 4$

Add 4 to both sides:

$a = 4 + 4$

$a = 8$

The first term is 8.

Now we need to find which term ($n$) of this AP is zero. Let $a_n = 0$.

Using the formula $a_n = a + (n-1)d$ with $a_n=0$, $a=8$, and $d=-2$:

$0 = 8 + (n - 1)(-2)$

$0 = 8 - 2(n - 1)$

Subtract 8 from both sides:

$-8 = -2(n - 1)$

Divide both sides by -2:

$\frac{-8}{-2} = n - 1$

$4 = n - 1$

Add 1 to both sides:

$n = 4 + 1$

$n = 5$

Since $n = 5$ is a positive integer, the 5th term of the AP is zero.

Thus, the 5th term of the AP is zero.

Given:

In an Arithmetic Progression (AP), let the first term be $a$ and the common difference be $d$.

The 17th term of the AP exceeds its 10th term by 7.

---

To Find:

The common difference ($d$) of the AP.

---

Solution:

The formula for the $n$-th term of an AP is given by:

$a_n = a + (n-1)d$

Using this formula, we can write the 17th term ($a_{17}$) and the 10th term ($a_{10}$) as follows:

$a_{17} = a + (17-1)d = a + 16d$

$a_{10} = a + (10-1)d = a + 9d$

According to the problem, the 17th term exceeds the 10th term by 7.

So, we have the equation:

$a_{17} - a_{10} = 7$

Substitute the expressions for $a_{17}$ and $a_{10}$ into this equation:

$(a + 16d) - (a + 9d) = 7$

Remove the parentheses:

$a + 16d - a - 9d = 7$

Combine like terms ($a$ terms cancel out):

$(a - a) + (16d - 9d) = 7$

$0 + 7d = 7$

$7d = 7$

Divide both sides by 7 to find the value of $d$:

$d = \frac{7}{7}$

$d = 1$

Thus, the common difference of the AP is 1.

Given:

The given Arithmetic Progression (AP) is 3, 15, 27, 39, . . .

First term, $a = 3$.

Common difference, $d = 15 - 3 = 12$.

---

To Find:

The term number ($n$) of the AP which is 132 more than its 54th term.

---

Solution:

The $n$-th term of an AP is given by the formula:

$a_n = a + (n-1)d$

First term $a = 3$.

Common difference $d = 12$.

We first find the 54th term of the AP.

$a_{54} = a + (54-1)d$

$a_{54} = 3 + (53) \times 12$

$a_{54} = 3 + 636$

$a_{54} = 639$

Let the required term be $a_n$. According to the problem, $a_n$ is 132 more than the 54th term.

$a_n = a_{54} + 132$

$a_n = 639 + 132$

$a_n = 771$

Now we need to find the term number ($n$) for which $a_n = 771$.

Using the formula for the $n$-th term:

$a_n = a + (n-1)d$

$771 = 3 + (n-1)12$

Subtract 3 from both sides:

$771 - 3 = (n-1)12$

$768 = (n-1)12$

Divide both sides by 12:

$n-1 = \frac{768}{12}$

$n-1 = 64$

Add 1 to both sides:

$n = 64 + 1$

$n = 65$

Therefore, the 65th term of the AP will be 132 more than its 54th term.

Given:

Let the first Arithmetic Progression (AP) have its first term as $A_1$ and the common difference as $d$.

Let the second Arithmetic Progression (AP) have its first term as $A_2$ and the common difference as $d$.

The common difference is the same for both APs.

The difference between their 100th terms is 100.

---

To Find:

The difference between their 1000th terms.

---

Solution:

The $n$-th term of an AP with first term $a$ and common difference $d$ is given by $a_n = a + (n-1)d$.

For the first AP, the 100th term is:

$a_{100} = A_1 + (100-1)d = A_1 + 99d$

For the second AP, the 100th term is:

$b_{100} = A_2 + (100-1)d = A_2 + 99d$

According to the problem, the difference between their 100th terms is 100.

So, $a_{100} - b_{100} = 100$

Substituting the expressions for $a_{100}$ and $b_{100}$:

$(A_1 + 99d) - (A_2 + 99d) = 100$

$A_1 + 99d - A_2 - 99d = 100$

$A_1 - A_2 = 100$

This equation shows that the difference between the first terms of the two APs is 100.

Now, let's find the 1000th term for both APs.

For the first AP, the 1000th term is:

$a_{1000} = A_1 + (1000-1)d = A_1 + 999d$

For the second AP, the 1000th term is:

$b_{1000} = A_2 + (1000-1)d = A_2 + 999d$

Now, we find the difference between their 1000th terms:

$a_{1000} - b_{1000} = (A_1 + 999d) - (A_2 + 999d)$

$a_{1000} - b_{1000} = A_1 + 999d - A_2 - 999d$

$a_{1000} - b_{1000} = A_1 - A_2$

We already found that $A_1 - A_2 = 100$.

Therefore, the difference between their 1000th terms is:

$a_{1000} - b_{1000} = 100$

The difference between the 1000th terms of the two APs is 100.

Given:

We need to find the number of three-digit numbers that are divisible by 7.

Three-digit numbers range from 100 to 999.

---

To Find:

The count of three-digit numbers divisible by 7.

---

Solution:

The three-digit numbers start from 100 and end at 999.

We need to find the smallest three-digit number divisible by 7 and the largest three-digit number divisible by 7.

To find the smallest three-digit number divisible by 7, we divide 100 by 7:

$100 \div 7 = 14$ with a remainder of 2.

So, the first multiple of 7 after 100 is $100 + (7 - 2) = 105$. This is the first term of our AP ($a$).

To find the largest three-digit number divisible by 7, we divide 999 by 7:

$999 \div 7 = 142$ with a remainder of 5.

So, the largest multiple of 7 before or at 999 is $999 - 5 = 994$. This is the last term of our AP ($a_n$).

The sequence of three-digit numbers divisible by 7 forms an Arithmetic Progression (AP):

105, 112, 119, . . . , 994

Here,

First term, $a = 105$

Common difference, $d = 7$ (since the numbers are divisible by 7)

Last term, $a_n = 994$

We use the formula for the $n$-th term of an AP:

$a_n = a + (n-1)d$

Substitute the values:

$994 = 105 + (n-1)7$

Subtract 105 from both sides:

$994 - 105 = (n-1)7$

$889 = (n-1)7$

Divide both sides by 7:

$\frac{889}{7} = n-1$

$127 = n-1$

Add 1 to both sides:

$n = 127 + 1$

$n = 128$

There are 128 three-digit numbers that are divisible by 7.

Given:

We need to find the number of multiples of 4 that lie between 10 and 250.

The numbers are in the range $(10, 250)$, i.e., from 11 to 249.

---

To Find:

The count of multiples of 4 between 10 and 250.

---

Solution:

The multiples of 4 between 10 and 250 form an Arithmetic Progression (AP).

We need to find the smallest multiple of 4 greater than 10.

Dividing 10 by 4, we get $10 = 4 \times 2 + 2$.

The smallest multiple of 4 greater than 10 is $10 + (4 - 2) = 12$.

So, the first term of the AP is $a = 12$.

We need to find the largest multiple of 4 less than 250.

Dividing 250 by 4, we get $250 = 4 \times 62 + 2$.

The largest multiple of 4 less than 250 is $250 - 2 = 248$.

So, the last term of the AP is $a_n = 248$.

The common difference is $d = 4$ as we are considering multiples of 4.

The AP is 12, 16, 20, . . . , 248.

To find the number of terms ($n$) in this AP, we use the formula for the $n$-th term:

$a_n = a + (n-1)d$

Substitute the values $a_n = 248$, $a = 12$, and $d = 4$:

$248 = 12 + (n-1)4$

Subtract 12 from both sides:

$248 - 12 = (n-1)4$

$236 = (n-1)4$

Divide both sides by 4:

$\frac{236}{4} = n-1$

$59 = n-1$

Add 1 to both sides:

$n = 59 + 1$

$n = 60$

Therefore, there are 60 multiples of 4 that lie between 10 and 250.

Given:

The first Arithmetic Progression (AP) is: 63, 65, 67, . . .

The second Arithmetic Progression (AP) is: 3, 10, 17, . . .

---

To Find:

The value of $n$ for which the $n$-th terms of the two APs are equal.

---

Solution:

For the first AP: 63, 65, 67, . . .

First term, $a_1 = 63$

Common difference, $d_1 = 65 - 63 = 2$

The $n$-th term of the first AP is given by:

$a_{1,n} = a_1 + (n-1)d_1 = 63 + (n-1)2$

For the second AP: 3, 10, 17, . . .

First term, $a_2 = 3$

Common difference, $d_2 = 10 - 3 = 7$

The $n$-th term of the second AP is given by:

$a_{2,n} = a_2 + (n-1)d_2 = 3 + (n-1)7$

We are looking for the value of $n$ where the $n$-th terms of the two APs are equal.

$a_{1,n} = a_{2,n}$

Substitute the expressions for the $n$-th terms:

$63 + (n-1)2 = 3 + (n-1)7$

Expand both sides of the equation:

$63 + 2n - 2 = 3 + 7n - 7$

Simplify both sides:

$61 + 2n = 7n - 4$

Rearrange the terms to group $n$ terms on one side and constant terms on the other side:

$61 + 4 = 7n - 2n$

$65 = 5n$

Solve for $n$ by dividing both sides by 5:

$n = \frac{65}{5}$

$n = 13$

Thus, the 13th term of both APs will be equal.

We can verify this by calculating the 13th term for each AP:

For the first AP: $a_{1,13} = 63 + (13-1)2 = 63 + 12 \times 2 = 63 + 24 = 87$

For the second AP: $a_{2,13} = 3 + (13-1)7 = 3 + 12 \times 7 = 3 + 84 = 87$

Since $a_{1,13} = a_{2,13} = 87$, the value of $n=13$ is correct.

Given:

In an Arithmetic Progression (AP), let the first term be $a$ and the common difference be $d$.

The third term ($a_3$) is 16.

The 7th term ($a_7$) exceeds the 5th term ($a_5$) by 12, which means $a_7 - a_5 = 12$.

---

To Find:

The Arithmetic Progression (AP).

---

Solution:

The formula for the $n$-th term of an AP is given by:

$a_n = a + (n-1)d$

Using the given information about the third term:

$a_3 = a + (3-1)d$

$16 = a + 2d$

This gives us the first equation:

$a + 2d = 16$

Using the information about the 7th and 5th terms:

$a_7 = a + (7-1)d = a + 6d$

$a_5 = a + (5-1)d = a + 4d$

We are given that $a_7 - a_5 = 12$.

Substitute the expressions for $a_7$ and $a_5$:

$(a + 6d) - (a + 4d) = 12$

Remove the parentheses:

$a + 6d - a - 4d = 12$

Combine like terms:

$(a - a) + (6d - 4d) = 12$

$0 + 2d = 12$

$2d = 12$

Divide by 2 to find the common difference $d$:

$d = \frac{12}{2}$

$d = 6$

Now substitute the value of $d = 6$ into the first equation ($a + 2d = 16$) to find the first term $a$:

$a + 2(6) = 16$

$a + 12 = 16$

Subtract 12 from both sides:

$a = 16 - 12$

$a = 4$

So, the first term is $a = 4$ and the common difference is $d = 6$.

The AP is formed by adding the common difference successively starting from the first term.

The terms are $a, a+d, a+2d, a+3d, \dots$

The AP is: $4, 4+6, 4+2(6), 4+3(6), \dots$

The AP is: $4, 10, 16, 22, \dots$

The AP is 4, 10, 16, 22, . . .

Given:

The given Arithmetic Progression (AP) is: 3, 8, 13, . . ., 253.

First term, $a = 3$.

Common difference, $d = 8 - 3 = 5$.

Last term, $l = 253$.

---

To Find:

The 20th term from the last term of the given AP.

---

Solution:

To find the 20th term from the last, we can consider the given AP in reverse order.

The reversed AP is: 253, . . ., 13, 8, 3.

For this reversed AP:

The first term is $a' = 253$.

The common difference is $d' = -d = -5$ (since the terms decrease by 5 in the reversed order).

We need to find the 20th term of this reversed AP. Let this term be $a'_{20}$.

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

Using this formula for the reversed AP (with first term $a'$ and common difference $d'$), the 20th term is:

$a'_{20} = a' + (20-1)d'$

$a'_{20} = 253 + (19)(-5)$

$a'_{20} = 253 - 95$

Now, perform the subtraction:

$a'_{20} = 158$

Therefore, the 20th term from the last term of the given AP is 158.

---

Alternate Solution:

We can also find the total number of terms ($n$) in the original AP first.

Using the formula $a_n = a + (n-1)d$ with $a_n = 253$, $a = 3$, and $d = 5$:

$253 = 3 + (n-1)5$

$250 = (n-1)5$

$50 = n-1$

$n = 51$

The total number of terms in the AP is 51.

The 20th term from the last is the $(51 - 20 + 1)$-th term from the beginning.

Term number from the beginning = $51 - 20 + 1 = 32$.

So, we need to find the 32nd term of the original AP.

$a_{32} = a + (32-1)d$

$a_{32} = 3 + (31)(5)$

$a_{32} = 3 + 155$

$a_{32} = 158$

Both methods yield the same result. The 20th term from the last of the AP is 158.

Given:

In an Arithmetic Progression (AP), let the first term be $a$ and the common difference be $d$.

The sum of the 4th term and the 8th term is 24 ($a_4 + a_8 = 24$).

The sum of the 6th term and the 10th term is 44 ($a_6 + a_{10} = 44$).

---

To Find:

The first three terms of the AP.

---

Solution:

The $n$-th term of an AP is given by the formula $a_n = a + (n-1)d$.

Using this formula, we can write the given information as equations:

The 4th term is $a_4 = a + (4-1)d = a + 3d$.

The 8th term is $a_8 = a + (8-1)d = a + 7d$.

The sum of the 4th and 8th terms is 24:

$a_4 + a_8 = 24$

$(a + 3d) + (a + 7d) = 24$

$2a + 10d = 24$

Divide the equation by 2:

The 6th term is $a_6 = a + (6-1)d = a + 5d$.

The 10th term is $a_{10} = a + (10-1)d = a + 9d$.

The sum of the 6th and 10th terms is 44:

$a_6 + a_{10} = 44$

$(a + 5d) + (a + 9d) = 44$

$2a + 14d = 44$

Divide the equation by 2:

Now we have a system of two linear equations with two variables $a$ and $d$:

$a + 5d = 12$ (1)

$a + 7d = 22$ (2)

Subtract equation (1) from equation (2):

$(a + 7d) - (a + 5d) = 22 - 12$

$a + 7d - a - 5d = 10$

$2d = 10$

Divide by 2:

$d = \frac{10}{2}$

$d = 5$

Substitute the value of $d = 5$ into equation (1):

$a + 5(5) = 12$

$a + 25 = 12$

Subtract 25 from both sides:

$a = 12 - 25$

$a = -13$

The first term of the AP is $a = -13$ and the common difference is $d = 5$.

The first three terms of the AP are:

First term: $a = -13$

Second term: $a + d = -13 + 5 = -8$

Third term: $a + 2d = -13 + 2(5) = -13 + 10 = -3$

The first three terms of the AP are -13, -8, and -3.

Given:

Subba Rao's starting annual salary in 1995 is $\textsf{₹} 5000$.

Annual increment is $\textsf{₹} 200$.

Target annual income is $\textsf{₹} 7000$.

---

To Find:

The year in which his income reached $\textsf{₹} 7000$.

---

Solution:

The annual salaries received by Subba Rao form an Arithmetic Progression (AP).

The salary in the first year (1995) is the first term, $a = \textsf{₹} 5000$.

The annual increment is the common difference, $d = \textsf{₹} 200$.

Let the year in which his income reached $\textsf{₹} 7000$ be the $n$-th year of his service.

The income in the $n$-th year is the $n$-th term of the AP, $a_n = \textsf{₹} 7000$.

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

Substitute the given values into the formula:

$7000 = 5000 + (n-1)200$

Subtract 5000 from both sides of the equation:

$7000 - 5000 = (n-1)200$

$2000 = (n-1)200$

Divide both sides by 200:

$\frac{2000}{200} = n-1$

$10 = n-1$

Add 1 to both sides to solve for $n$:

$n = 10 + 1$

$n = 11$

So, the 11th term of the AP corresponds to the income of $\textsf{₹} 7000$.

The first term corresponds to the salary in the year 1995.

The $n$-th term corresponds to the salary in the year $1995 + (n-1)$.

For $n = 11$, the year is $1995 + (11-1) = 1995 + 10 = 2005$.

Therefore, Subba Rao's income reached $\textsf{₹} 7000$ in the year 2005.

Given:

Ramkali's saving in the first week = $\textsf{₹} 5$.

Increase in weekly savings = $\textsf{₹} 1.75$ per week.

Her weekly saving in the $n$-th week = $\textsf{₹} 20.75$.

---

To Find:

The value of $n$ (the week number) when her weekly savings become $\textsf{₹} 20.75$.

---

Solution:

The weekly savings of Ramkali form an Arithmetic Progression (AP).

The saving in the first week is the first term, $a = \textsf{₹} 5$.

The increase in weekly saving is the common difference, $d = \textsf{₹} 1.75$.

The weekly saving in the $n$-th week is the $n$-th term of the AP, $a_n = \textsf{₹} 20.75$.

The formula for the $n$-th term of an AP is given by:

$a_n = a + (n-1)d$

Substitute the given values into the formula:

$20.75 = 5 + (n-1)1.75$

Subtract 5 from both sides of the equation:

$20.75 - 5 = (n-1)1.75$

$15.75 = (n-1)1.75$

Divide both sides by 1.75:

$\frac{15.75}{1.75} = n-1$

To divide 15.75 by 1.75, we can remove the decimal points by multiplying both numerator and denominator by 100:

$\frac{1575}{175} = n-1$

Simplify the fraction:

$\frac{\cancel{1575}^{63}}{\cancel{175}_{7}} = n-1$ (Dividing by 25)

$\frac{63}{7} = n-1$

$9 = n-1$

Add 1 to both sides to solve for $n$:

$n = 9 + 1$

$n = 10$

Therefore, in the 10th week, Ramkali's weekly savings will become $\textsf{₹} 20.75$.

Given:

The given Arithmetic Progression (AP) is: 8, 3, –2, . . .

We need to find the sum of the first 22 terms of this AP.

---

To Find:

The sum of the first 22 terms ($S_{22}$) of the AP.

---

Solution:

From the given AP, we can identify the first term and the common difference:

First term, $a = 8$

Common difference, $d = 3 - 8 = -5$

The number of terms we need to sum is $n = 22$.

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values of $n$, $a$, and $d$ into the formula:

$S_{22} = \frac{22}{2}[2(8) + (22-1)(-5)]$

$S_{22} = 11[16 + (21)(-5)]$

$S_{22} = 11[16 - 105]$

$S_{22} = 11[-89]$

Now, perform the multiplication:

$S_{22} = -979$

The sum of the first 22 terms of the AP is -979.

Given:

The sum of the first 14 terms of an AP is 1050 ($S_{14} = 1050$).

The first term is 10 ($a = 10$).

---

To Find:

The 20th term ($a_{20}$) of the AP.

---

Solution:

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

We are given $S_{14} = 1050$, $n = 14$, and $a = 10$. We can use this information to find the common difference $d$.

Substitute the given values into the sum formula:

$1050 = \frac{14}{2}[2(10) + (14-1)d]$

$1050 = 7[20 + 13d]$

Divide both sides of the equation by 7:

$\frac{1050}{7} = 20 + 13d$

$150 = 20 + 13d$

Subtract 20 from both sides:

$150 - 20 = 13d$

$130 = 13d$

Divide both sides by 13:

$d = \frac{130}{13}$

$d = 10$

Now that we have the common difference $d = 10$, we can find the 20th term using the formula for the $n$-th term of an AP:

$a_n = a + (n-1)d$

We want to find the 20th term ($a_{20}$), so we set $n = 20$, $a = 10$, and $d = 10$.

$a_{20} = 10 + (20-1)10$

$a_{20} = 10 + (19)10$

$a_{20} = 10 + 190$

$a_{20} = 200$

The 20th term of the AP is 200.

Given:

The given Arithmetic Progression (AP) is: 24, 21, 18, . . .

The sum of $n$ terms ($S_n$) is 78.

---

To Find:

The number of terms ($n$) that must be taken so that their sum is 78.

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Solution:

From the given AP, we have:

First term, $a = 24$

Common difference, $d = 21 - 24 = -3$

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2}[2a + (n-1)d]$

We are given that the sum $S_n = 78$. Substitute the values of $S_n$, $a$, and $d$ into the formula:

$78 = \frac{n}{2}[2(24) + (n-1)(-3)]$

Multiply both sides by 2:

$2 \times 78 = n[48 - 3(n-1)]$

$156 = n[48 - 3n + 3]$

$156 = n[51 - 3n]$

$156 = 51n - 3n^2$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

$3n^2 - 51n + 156 = 0$

Divide the entire equation by 3 to simplify:

$\frac{3n^2}{3} - \frac{51n}{3} + \frac{156}{3} = \frac{0}{3}$

$n^2 - 17n + 52 = 0$

Now, we solve this quadratic equation for $n$ by factoring or using the quadratic formula.

We look for two numbers that multiply to 52 and add up to -17. These numbers are -4 and -13.

So, we can factor the quadratic equation as:

$(n - 4)(n - 13) = 0$

This gives two possible values for $n$:

$n - 4 = 0 \implies n = 4$

or

$n - 13 = 0 \implies n = 13$

Both $n=4$ and $n=13$ are positive integers, which are valid numbers of terms for an AP.

Let's check the sum for both values of $n$:

If $n = 4$:

The terms are 24, 21, 18, 15. Their sum is $24 + 21 + 18 + 15 = 78$. This is correct.

If $n = 13$:

The terms of the AP are 24, 21, 18, 15, 12, 9, 6, 3, 0, -3, -6, -9, -12.

The sum of the first 4 terms is 78. The terms from the 5th to the 13th are 12, 9, 6, 3, 0, -3, -6, -9, -12. The sum of these terms is $(12 + (-12)) + (9 + (-9)) + (6 + (-6)) + (3 + (-3)) + 0 = 0 + 0 + 0 + 0 + 0 = 0$.

So, the sum of the first 13 terms is the sum of the first 4 terms plus the sum of the terms from the 5th to 13th, which is $78 + 0 = 78$. This is also correct.

Since the common difference is negative, the terms become smaller and eventually turn negative. The sum increases up to a certain point (where the terms are positive) and then starts decreasing as negative terms are added. In this case, the sum is 78 after 4 terms and again after 13 terms because the sum of the terms from the 5th to the 13th is zero.

Thus, the number of terms that must be taken is 4 or 13.

Given:

(i) The sequence of the first 1000 positive integers.

(ii) The sequence of the first $n$ positive integers.

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To Find:

(i) The sum of the first 1000 positive integers.

(ii) The sum of the first $n$ positive integers.

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Solution:

The sequence of positive integers 1, 2, 3, 4, . . . forms an Arithmetic Progression (AP).

The first term is $a = 1$.

The common difference is $d = 2 - 1 = 1$.

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + a_n)$, where $a_n$ is the $n$-th term.

---

(i) Sum of the first 1000 positive integers:

Here, the number of terms is $n = 1000$.

The first term is $a = 1$.

The 1000th term is $a_{1000} = 1000$ (since the common difference is 1).

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{1000} = \frac{1000}{2}(1 + 1000)$

$S_{1000} = 500(1001)$

$S_{1000} = 500500$

The sum of the first 1000 positive integers is 500500.

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(ii) Sum of the first n positive integers:

Here, the number of terms is $n$.

The first term is $a = 1$.

The common difference is $d = 1$.

The $n$-th term is $a_n = a + (n-1)d = 1 + (n-1)(1) = 1 + n - 1 = n$.

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$:

$S_n = \frac{n}{2}(1 + n)$

$S_n = \frac{n(n+1)}{2}$

The sum of the first $n$ positive integers is given by the formula:

$S_n = \frac{n(n+1)}{2}$

Given:

The $n$-th term of a list of numbers is given by $a_n = 3 + 2n$.

We need to find the sum of the first 24 terms ($n=24$).

---

To Find:

The sum of the first 24 terms ($S_{24}$).

---

Solution:

The given formula for the $n$-th term is $a_n = 3 + 2n$.

Let's find the first few terms of the list to see if it is an AP.

For $n=1$, the first term is $a_1 = 3 + 2(1) = 3 + 2 = 5$.

For $n=2$, the second term is $a_2 = 3 + 2(2) = 3 + 4 = 7$.

For $n=3$, the third term is $a_3 = 3 + 2(3) = 3 + 6 = 9$.

The list of numbers is 5, 7, 9, . . .

The difference between consecutive terms is $a_2 - a_1 = 7 - 5 = 2$ and $a_3 - a_2 = 9 - 7 = 2$.

Since the difference between consecutive terms is constant, this list of numbers forms an Arithmetic Progression (AP) with:

First term, $a = 5$

Common difference, $d = 2$

We need to find the sum of the first 24 terms, so $n = 24$.

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Alternatively, we can find the 24th term first and then use the formula $S_n = \frac{n}{2}(a + a_n)$.

Using the given formula $a_n = 3 + 2n$, the 24th term is:

$a_{24} = 3 + 2(24) = 3 + 48 = 51$.

Now, using the sum formula $S_n = \frac{n}{2}(a + a_n)$ with $n=24$, $a=5$, and $a_{24}=51$:

$S_{24} = \frac{24}{2}(5 + 51)$

$S_{24} = 12(56)$

Now perform the multiplication:

$S_{24} = 672$

The sum of the first 24 terms of the list is 672.

Given:

The production of TV sets increases uniformly each year, which means the annual production figures form an Arithmetic Progression (AP).

Let the production in the $n$-th year be the $n$-th term of the AP, denoted by $a_n$.

Production in the 3rd year ($a_3$) = 600 sets.

Production in the 7th year ($a_7$) = 700 sets.

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To Find:

(i) The production in the 1st year ($a$).

(ii) The production in the 10th year ($a_{10}$).

(iii) The total production in the first 7 years ($S_7$).

---

Solution:

Let $a$ be the production in the first year and $d$ be the uniform annual increase (common difference).

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

Using the given information:

The 3rd term is $a_3 = a + (3-1)d = a + 2d$. So, we have:

The 7th term is $a_7 = a + (7-1)d = a + 6d$. So, we have:

Now we have a system of two linear equations. Subtract equation (1) from equation (2) to find $d$:

$(a + 6d) - (a + 2d) = 700 - 600$

$a + 6d - a - 2d = 100$

$4d = 100$

Divide by 4:

$d = \frac{100}{4}$

$d = 25$

Substitute the value of $d = 25$ into equation (1) to find $a$:

$a + 2(25) = 600$

$a + 50 = 600$

$a = 600 - 50$

$a = 550$

So, the production in the 1st year is 550 sets, and the uniform annual increase is 25 sets.

---

(i) Production in the 1st year:

The production in the 1st year is the first term of the AP, $a$.

$a = 550$ sets.

---

(ii) Production in the 10th year:

We need to find the 10th term of the AP, $a_{10}$.

Using the formula $a_n = a + (n-1)d$ with $n=10$, $a=550$, and $d=25$:

$a_{10} = 550 + (10-1)25$

$a_{10} = 550 + 9(25)$

$a_{10} = 550 + 225$

$a_{10} = 775$

The production in the 10th year is 775 sets.

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(iii) Total production in first 7 years:

We need to find the sum of the first 7 terms of the AP, $S_7$.

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Substitute $n=7$, $a=550$, and $d=25$ into the formula:

$S_7 = \frac{7}{2}[2(550) + (7-1)25]$

$S_7 = \frac{7}{2}[1100 + 6(25)]$

$S_7 = \frac{7}{2}[1100 + 150]$

$S_7 = \frac{7}{2}[1250]$

$S_7 = 7 \times \frac{1250}{2}$

$S_7 = 7 \times 625$

To calculate $7 \times 625$:

$7 \times 625 = 7 \times (600 + 20 + 5)$

$= 7 \times 600 + 7 \times 20 + 7 \times 5$

$= 4200 + 140 + 35$

$= 4375$

$S_7 = 4375$

The total production in the first 7 years is 4375 sets.

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference.

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(i) AP: 2, 7, 12, . . ., to 10 terms.

Given:

First term, $a = 2$.

Common difference, $d = 7 - 2 = 5$.

Number of terms, $n = 10$.

To Find:

Sum of the first 10 terms ($S_{10}$).

Solution:

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{10} = \frac{10}{2}[2(2) + (10-1)5]$

$S_{10} = 5[4 + (9)5]$

$S_{10} = 5[4 + 45]$

$S_{10} = 5[49]$

$S_{10} = 245$

The sum of the first 10 terms is 245.

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(ii) AP: –37, –33, –29, . . ., to 12 terms.

Given:

First term, $a = -37$.

Common difference, $d = -33 - (-37) = -33 + 37 = 4$.

Number of terms, $n = 12$.

To Find:

Sum of the first 12 terms ($S_{12}$).

Solution:

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{12} = \frac{12}{2}[2(-37) + (12-1)4]$

$S_{12} = 6[-74 + (11)4]$

$S_{12} = 6[-74 + 44]$

$S_{12} = 6[-30]$

$S_{12} = -180$

The sum of the first 12 terms is -180.

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(iii) AP: 0.6, 1.7, 2.8, . . ., to 100 terms.

Given:

First term, $a = 0.6$.

Common difference, $d = 1.7 - 0.6 = 1.1$.

Number of terms, $n = 100$.

To Find:

Sum of the first 100 terms ($S_{100}$).

Solution:

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]$

$S_{100} = 50[1.2 + (99)1.1]$

$S_{100} = 50[1.2 + 108.9]$

$S_{100} = 50[110.1]$

$S_{100} = 50 \times 110.1 = 5 \times 1101$

$S_{100} = 5505$

The sum of the first 100 terms is 5505.

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(iv) AP: $\frac{1}{15}$ , $\frac{1}{12}$ , $\frac{1}{10}$ , . . ., to 11 terms.

Given:

First term, $a = \frac{1}{15}$.

Common difference, $d = \frac{1}{12} - \frac{1}{15}$.

To find the common difference, find a common denominator for 12 and 15, which is 60.

$d = \frac{5}{60} - \frac{4}{60} = \frac{5-4}{60} = \frac{1}{60}$.

Number of terms, $n = 11$.

To Find:

Sum of the first 11 terms ($S_{11}$).

Solution:

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{11} = \frac{11}{2}[2(\frac{1}{15}) + (11-1)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + (10)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{10}{60}]$

Simplify the fraction $\frac{10}{60} = \frac{1}{6}$.

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}]$

Find a common denominator for 15 and 6, which is 30.

$S_{11} = \frac{11}{2}[\frac{2 \times 2}{15 \times 2} + \frac{1 \times 5}{6 \times 5}]$

$S_{11} = \frac{11}{2}[\frac{4}{30} + \frac{5}{30}]$

$S_{11} = \frac{11}{2}[\frac{4+5}{30}]$

$S_{11} = \frac{11}{2}[\frac{9}{30}]$

Simplify the fraction $\frac{9}{30} = \frac{3}{10}$.

$S_{11} = \frac{11}{2} \times \frac{3}{10}$

$S_{11} = \frac{11 \times 3}{2 \times 10}$

$S_{11} = \frac{33}{20}$

The sum of the first 11 terms is $\frac{33}{20}$.

The sum of an AP can be found using the formula $S_n = \frac{n}{2}(a + a_n)$, where $a$ is the first term, $a_n$ is the last term, and $n$ is the number of terms.

To find $n$, we use the formula $a_n = a + (n-1)d$, where $d$ is the common difference.

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(i) 7 + $10\frac{1}{2}$ + 14 + . . . + 84

Given:

First term, $a = 7$.

Last term, $a_n = 84$.

Common difference, $d = 10\frac{1}{2} - 7 = \frac{21}{2} - \frac{14}{2} = \frac{7}{2}$.

To Find:

The sum of the given AP.

Solution:

First, find the number of terms $n$ using the formula $a_n = a + (n-1)d$:

$84 = 7 + (n-1)\frac{7}{2}$

$84 - 7 = (n-1)\frac{7}{2}$

$77 = (n-1)\frac{7}{2}$

Multiply both sides by $\frac{2}{7}$:

$77 \times \frac{2}{7} = n-1$

$\cancel{77}^{11} \times \frac{2}{\cancel{7}_{1}} = n-1$

$11 \times 2 = n-1$

$22 = n-1$

$n = 22 + 1 = 23$

Now, find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{23} = \frac{23}{2}(7 + 84)$

$S_{23} = \frac{23}{2}(91)$

$S_{23} = \frac{2093}{2}$

$S_{23} = 1046.5$

The sum of the given AP is $\frac{2093}{2}$ or $1046.5$.

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(ii) 34 + 32 + 30 + . . . + 10

Given:

First term, $a = 34$.

Last term, $a_n = 10$.

Common difference, $d = 32 - 34 = -2$.

To Find:

The sum of the given AP.

Solution:

First, find the number of terms $n$ using the formula $a_n = a + (n-1)d$:

$10 = 34 + (n-1)(-2)$

$10 - 34 = (n-1)(-2)$

$-24 = -2(n-1)$

Divide both sides by -2:

$\frac{-24}{-2} = n-1$

$12 = n-1$

$n = 12 + 1 = 13$

Now, find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{13} = \frac{13}{2}(34 + 10)$

$S_{13} = \frac{13}{2}(44)$

$S_{13} = 13 \times \frac{44}{2}$

$S_{13} = 13 \times 22$

$S_{13} = 286$

The sum of the given AP is 286.

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(iii) –5 + (–8) + (–11) + . . . + (–230)

Given:

First term, $a = -5$.

Last term, $a_n = -230$.

Common difference, $d = -8 - (-5) = -8 + 5 = -3$.

To Find:

The sum of the given AP.

Solution:

First, find the number of terms $n$ using the formula $a_n = a + (n-1)d$:

$-230 = -5 + (n-1)(-3)$

$-230 + 5 = (n-1)(-3)$

$-225 = -3(n-1)$

Divide both sides by -3:

$\frac{-225}{-3} = n-1$

$75 = n-1$

$n = 75 + 1 = 76$

Now, find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{76} = \frac{76}{2}(-5 + (-230))$

$S_{76} = 38(-235)$

$S_{76} = -8930$

The sum of the given AP is -8930.

The formulas used are the $n$-th term of an AP, $a_n = a + (n-1)d$, and the sum of the first $n$ terms, $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + a_n)$.

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(i) Given a = 5, d = 3, a_n = 50, find n and S_n.

Given: $a=5$, $d=3$, $a_n=50$.

To Find: $n$, $S_n$.

Using the formula for the $n$-th term:

$a_n = a + (n-1)d$

$50 = 5 + (n-1)3$

$50 - 5 = (n-1)3$

$45 = 3(n-1)$

$\frac{45}{3} = n-1$

$15 = n-1$

$n = 15 + 1 = 16$

Now find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{16} = \frac{16}{2}(5 + 50)$

$S_{16} = 8(55)$

$S_{16} = 440$

Therefore, $n=16$ and $S_n=440$.

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(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.

Given: $a=7$, $a_{13}=35$, $n=13$.

To Find: $d$, $S_{13}$.

Using the formula for the $n$-th term ($n=13$):

$a_{13} = a + (13-1)d$

$35 = 7 + 12d$

$35 - 7 = 12d$

$28 = 12d$

$d = \frac{28}{12} = \frac{7}{3}$

Now find the sum $S_{13}$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{13} = \frac{13}{2}(7 + 35)$

$S_{13} = \frac{13}{2}(42)$

$S_{13} = 13 \times \frac{42}{2}$

$S_{13} = 13 \times 21$

$S_{13} = 273$

Therefore, $d=\frac{7}{3}$ and $S_{13}=273$.

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(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.

Given: $a_{12}=37$, $d=3$, $n=12.

To Find: $a$, $S_{12}$.

Using the formula for the $n$-th term ($n=12$):

$a_{12} = a + (12-1)d$

$37 = a + 11(3)

$37 = a + 33$

$a = 37 - 33 = 4$

Now find the sum $S_{12}$ using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{12} = \frac{12}{2}(4 + 37)$

$S_{12} = 6(41)$

$S_{12} = 246$

Therefore, $a=4$ and $S_{12}=246$.

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(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Given: $a_3=15$, $S_{10}=125$.

To Find: $d$, $a_{10}$.

Using the formula for the $n$-th term for $n=3$:

$a_3 = a + (3-1)d$

Using the formula for the sum of the first 10 terms ($n=10$):

$S_{10} = \frac{10}{2}[2a + (10-1)d]$

$125 = 5[2a + 9d]$

Divide by 5:

$\frac{125}{5} = 2a + 9d$

From equation (1), we can express $a$ as $a = 15 - 2d$. Substitute this into equation (2):

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

$25 - 30 = 5d$

$-5 = 5d$

$d = \frac{-5}{5} = -1$

Now substitute the value of $d = -1$ back into equation (1) to find $a$:

$15 = a + 2(-1)$

$15 = a - 2$

$a = 15 + 2 = 17$

Now find the 10th term ($a_{10}$) using the formula $a_n = a + (n-1)d$:

$a_{10} = a + (10-1)d$

$a_{10} = 17 + 9(-1)$

$a_{10} = 17 - 9 = 8$

Therefore, $d=-1$ and $a_{10}=8$.

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(v) Given d = 5, S₉ = 75, find a and a₉.

Given: $d=5$, $S_9=75$, $n=9$.

To Find: $a$, $a_9$.

Using the formula for the sum of the first 9 terms ($n=9$):

$S_9 = \frac{9}{2}[2a + (9-1)d]$

$75 = \frac{9}{2}[2a + 8(5)]$

$75 = \frac{9}{2}[2a + 40]$

Multiply both sides by $\frac{2}{9}$:

$75 \times \frac{2}{9} = 2a + 40$

$\cancel{75}^{25} \times \frac{2}{\cancel{9}_{3}} = 2a + 40$

$\frac{50}{3} = 2a + 40$

Subtract 40 from both sides:

$\frac{50}{3} - 40 = 2a$

$\frac{50 - 40 \times 3}{3} = 2a$

$\frac{50 - 120}{3} = 2a$

$\frac{-70}{3} = 2a$

Divide by 2:

$a = \frac{-70}{3 \times 2} = \frac{-70}{6} = \frac{-35}{3}$

Now find the 9th term ($a_9$) using the formula $a_n = a + (n-1)d$:

$a_9 = a + (9-1)d$

$a_9 = \frac{-35}{3} + 8(5)$

$a_9 = \frac{-35}{3} + 40$

$a_9 = \frac{-35 + 40 \times 3}{3}$

$a_9 = \frac{-35 + 120}{3} = \frac{85}{3}$

Therefore, $a=\frac{-35}{3}$ and $a_9=\frac{85}{3}$.

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(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.

Given: $a=2$, $d=8$, $S_n=90$.

To Find: $n$, $a_n$.

Using the formula for the sum of the first $n$ terms:

$S_n = \frac{n}{2}[2a + (n-1)d]$

$90 = \frac{n}{2}[2(2) + (n-1)8]$

$90 = \frac{n}{2}[4 + 8n - 8]$

$90 = \frac{n}{2}[8n - 4]$

Multiply both sides by 2:

$180 = n(8n - 4)$

$180 = 8n^2 - 4n$

Rearrange into a quadratic equation:

$8n^2 - 4n - 180 = 0$

Divide by 4:

$2n^2 - n - 45 = 0$

Solve the quadratic equation for $n$. We can factor or use the quadratic formula. Factoring: We need two numbers that multiply to $2 \times (-45) = -90$ and add to -1. These numbers are -10 and 9.

$2n^2 - 10n + 9n - 45 = 0$

$2n(n - 5) + 9(n - 5) = 0$

$(n - 5)(2n + 9) = 0$

Setting each factor to zero:

$n - 5 = 0 \implies n = 5$

$2n + 9 = 0 \implies 2n = -9 \implies n = -\frac{9}{2}$

Since the number of terms must be a positive integer, we take $n=5$.

Now find the $n$-th term ($a_n$) for $n=5$, i.e., $a_5$, using the formula $a_n = a + (n-1)d$:

$a_5 = 2 + (5-1)8$

$a_5 = 2 + 4(8)

$a_5 = 2 + 32 = 34$

Therefore, $n=5$ and $a_n=34$.

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(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.

Given: $a=8$, $a_n=62$, $S_n=210$.

To Find: $n$, $d$.

Using the formula for the sum of the first $n$ terms $S_n = \frac{n}{2}(a + a_n)$:

$210 = \frac{n}{2}(8 + 62)

$210 = \frac{n}{2}(70)

$210 = 35n$

$n = \frac{210}{35}$

$n = 6$

Now use the formula for the $n$-th term $a_n = a + (n-1)d$ with $n=6$:

$a_6 = a + (6-1)d$

$62 = 8 + 5d$

$62 - 8 = 5d$

$54 = 5d$

$d = \frac{54}{5}$

Therefore, $n=6$ and $d=\frac{54}{5}$.

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(viii) Given a_n = 4, d = 2, S_n = –14, find n and a.

Given: $a_n=4$, $d=2$, $S_n=-14$.

To Find: $n$, $a$.

Using the formula for the $n$-th term $a_n = a + (n-1)d$:

$4 = a + (n-1)2$

$4 = a + 2n - 2$

$a = 4 - 2n + 2$

Using the formula for the sum of the first $n$ terms $S_n = \frac{n}{2}(a + a_n)$, and substitute the expression for $a$ from (1):

$-14 = \frac{n}{2}((6 - 2n) + 4)$

$-14 = \frac{n}{2}(10 - 2n)$

Multiply both sides by 2:

$-28 = n(10 - 2n)$

$-28 = 10n - 2n^2$

Rearrange into a quadratic equation:

$2n^2 - 10n - 28 = 0$

Divide by 2:

$n^2 - 5n - 14 = 0$

Solve the quadratic equation for $n$ by factoring. We need two numbers that multiply to -14 and add to -5. These numbers are -7 and 2.

$(n - 7)(n + 2) = 0$

Setting each factor to zero:

$n - 7 = 0 \implies n = 7$

$n + 2 = 0 \implies n = -2$

Since the number of terms must be a positive integer, we take $n=7$.

Now find the first term $a$ using equation (1) with $n=7$:

$a = 6 - 2(7)$

$a = 6 - 14 = -8$

Therefore, $n=7$ and $a=-8$.

---

(ix) Given a = 3, n = 8, S = 192, find d.

Given: $a=3$, $n=8$, $S_8=192$.

To Find: $d$.

Using the formula for the sum of the first $n$ terms:

$S_n = \frac{n}{2}[2a + (n-1)d]$

$S_8 = \frac{8}{2}[2(3) + (8-1)d]$

$192 = 4[6 + 7d]$

Divide both sides by 4:

$\frac{192}{4} = 6 + 7d$

$48 = 6 + 7d$

$48 - 6 = 7d$

$42 = 7d$

$d = \frac{42}{7}$

$d = 6$

Therefore, $d=6$.

---

(x) Given l = 28, S = 144, and there are total 9 terms. Find a.

Given: Last term $l = a_n = 28$, sum $S_n = 144$, number of terms $n = 9$.

To Find: First term $a$.

Using the formula for the sum of the first $n$ terms when the last term is known:

$S_n = \frac{n}{2}(a + a_n)$

$S_9 = \frac{9}{2}(a + a_9)$

$144 = \frac{9}{2}(a + 28)$

Multiply both sides by $\frac{2}{9}$:

$144 \times \frac{2}{9} = a + 28$

$\cancel{144}^{16} \times \frac{2}{\cancel{9}_{1}} = a + 28$

$16 \times 2 = a + 28$

$32 = a + 28$

$a = 32 - 28$

$a = 4$

Therefore, the first term $a=4$.

Given:

The Arithmetic Progression (AP) is: 9, 17, 25, . . .

The sum of $n$ terms ($S_n$) is 636.

---

To Find:

The number of terms ($n$) that must be taken to give a sum of 636.

---

Solution:

From the given AP, we have:

First term, $a = 9$.

Common difference, $d = 17 - 9 = 8$.

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2}[2a + (n-1)d]$

We are given that the sum $S_n = 636$. Substitute the values of $S_n$, $a$, and $d$ into the formula:

$636 = \frac{n}{2}[2(9) + (n-1)8]$

$636 = \frac{n}{2}[18 + 8n - 8]$

$636 = \frac{n}{2}[10 + 8n]$

Multiply both sides by 2:

$2 \times 636 = n(10 + 8n)$

$1272 = 10n + 8n^2$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

$8n^2 + 10n - 1272 = 0$

Divide the entire equation by 2 to simplify:

$\frac{8n^2}{2} + \frac{10n}{2} - \frac{1272}{2} = \frac{0}{2}$

$4n^2 + 5n - 636 = 0$

Now, we solve this quadratic equation for $n$ using the quadratic formula:

$n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Here, $A=4$, $B=5$, and $C=-636$.

$n = \frac{-5 \pm \sqrt{5^2 - 4(4)(-636)}}{2(4)}$

$n = \frac{-5 \pm \sqrt{25 - 16(-636)}}{8}$

$n = \frac{-5 \pm \sqrt{25 + 10176}}{8}$

$n = \frac{-5 \pm \sqrt{10201}}{8}$

Find the square root of 10201. The square root of 10201 is 101.

$n = \frac{-5 \pm 101}{8}$

We have two possible values for $n$:

$n_1 = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

$n_2 = \frac{-5 - 101}{8} = \frac{-106}{8} = -\frac{53}{4}$

Since the number of terms ($n$) must be a positive integer, we discard the negative and fractional value.

Therefore, $n = 12$.

12 terms of the AP must be taken to give a sum of 636.

Given:

First term of the AP, $a = 5$.

Last term of the AP, $a_n = 45$.

Sum of the AP, $S_n = 400$.

---

To Find:

The number of terms ($n$) and the common difference ($d$) of the AP.

---

Solution:

The formula for the sum of the first $n$ terms of an AP, when the first and last terms are known, is:

$S_n = \frac{n}{2}(a + a_n)$

Substitute the given values $S_n = 400$, $a = 5$, and $a_n = 45$ into this formula:

$400 = \frac{n}{2}(5 + 45)$

$400 = \frac{n}{2}(50)$

Simplify the right side:

$400 = 25n$

Solve for $n$ by dividing both sides by 25:

$n = \frac{400}{25}$

$n = 16$

So, there are 16 terms in the AP.

Now, we use the formula for the $n$-th term of an AP to find the common difference $d$:

$a_n = a + (n-1)d$

Substitute the values $a_n = 45$, $a = 5$, and $n = 16$ (which we just found):

$45 = 5 + (16-1)d$

$45 = 5 + 15d$

Subtract 5 from both sides:

$45 - 5 = 15d$

$40 = 15d$

Solve for $d$ by dividing both sides by 15:

$d = \frac{40}{15}$

Simplify the fraction:

$d = \frac{\cancel{40}^{8}}{\cancel{15}_{3}}$

$d = \frac{8}{3}$

The number of terms is 16 and the common difference is $\frac{8}{3}$.

Given:

First term of the AP, $a = 17$.

Last term of the AP, $a_n = 350$.

Common difference, $d = 9$.

---

To Find:

The number of terms ($n$) and the sum of the terms ($S_n$).

---

Solution:

We use the formula for the $n$-th term of an AP to find the number of terms $n$:

$a_n = a + (n-1)d$

Substitute the given values $a_n = 350$, $a = 17$, and $d = 9$ into the formula:

$350 = 17 + (n-1)9$

Subtract 17 from both sides:

$350 - 17 = (n-1)9$

$333 = (n-1)9$

Divide both sides by 9:

$\frac{333}{9} = n-1$

$37 = n-1$

Add 1 to both sides:

$n = 37 + 1$

$n = 38$

So, there are 38 terms in the AP.

Now, we find the sum of these 38 terms ($S_{38}$) using the formula for the sum of the first $n$ terms when the first and last terms are known:

$S_n = \frac{n}{2}(a + a_n)$

Substitute the values $n = 38$, $a = 17$, and $a_n = 350$ into this formula:

$S_{38} = \frac{38}{2}(17 + 350)$

$S_{38} = 19(367)$

$S_{38} = 6973$

There are 38 terms in the AP, and their sum is 6973.

Given:

Common difference of the AP, $d = 7$.

The 22nd term of the AP, $a_{22} = 149$.

We need to find the sum of the first 22 terms ($n=22$).

---

To Find:

The sum of the first 22 terms ($S_{22}$).

---

Solution:

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

We are given the 22nd term ($a_{22} = 149$) and the common difference ($d = 7$). We can use this to find the first term ($a$).

For $n = 22$:

$a_{22} = a + (22-1)d$

$149 = a + (21)(7)$

$149 = a + 147$

$a = 149 - 147$

$a = 2$

The first term of the AP is $a = 2$.

Now, we can find the sum of the first 22 terms ($S_{22}$). The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}(a + a_n)$, when the first and last terms are known.

Here, $n = 22$, the first term is $a = 2$, and the 22nd term (which is the last term in this sum) is $a_{22} = 149$.

Using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{22} = \frac{22}{2}(2 + 149)$

$S_{22} = 11(151)$

$S_{22} = 1661$

The sum of the first 22 terms of the AP is 1661.

Given:

In an Arithmetic Progression (AP):

Second term, $a_2 = 14$.

Third term, $a_3 = 18$.

We need to find the sum of the first 51 terms, so $n=51$.

---

To Find:

The sum of the first 51 terms ($S_{51}$) of the AP.

---

Solution:

Let the first term of the AP be $a$ and the common difference be $d$.

The terms of the AP are $a, a+d, a+2d, \dots, a+(n-1)d$.

The second term is given by $a_2 = a + (2-1)d = a + d$.

The third term is given by $a_3 = a + (3-1)d = a + 2d$.

To find the common difference $d$, subtract equation (1) from equation (2):

$(a + 2d) - (a + d) = 18 - 14$

$a + 2d - a - d = 4$

$d = 4$

Now, substitute the value of $d = 4$ into equation (1) to find the first term $a$:

$a + d = 14$

$a + 4 = 14$

$a = 14 - 4$

$a = 10$

So, the first term is $a=10$ and the common difference is $d=4$.

We need to find the sum of the first 51 terms ($S_{51}$). The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $n=51$, $a=10$, and $d=4$ into the formula:

$S_{51} = \frac{51}{2}[2(10) + (51-1)4]$

$S_{51} = \frac{51}{2}[20 + (50)4]$

$S_{51} = \frac{51}{2}[20 + 200]$

$S_{51} = \frac{51}{2}[220]$

Simplify the multiplication:

$S_{51} = 51 \times \frac{220}{2}$

$S_{51} = 51 \times 110$

$S_{51} = 5610$

The sum of the first 51 terms of the AP is 5610.

Given:

In an Arithmetic Progression (AP), let the first term be $a$ and the common difference be $d$.

The sum of the first 7 terms ($S_7$) is 49.

The sum of the first 17 terms ($S_{17}$) is 289.

---

To Find:

The sum of the first $n$ terms ($S_n$).

---

Solution:

The formula for the sum of the first $k$ terms of an AP is given by:

$S_k = \frac{k}{2}[2a + (k-1)d]$

Using the given information for $S_7 = 49$ and $n=7$:

$S_7 = \frac{7}{2}[2a + (7-1)d]$

$49 = \frac{7}{2}[2a + 6d]$

Multiply both sides by $\frac{2}{7}$:

$49 \times \frac{2}{7} = 2a + 6d$

$14 = 2a + 6d$

Divide the equation by 2:

Using the given information for $S_{17} = 289$ and $n=17$:

$S_{17} = \frac{17}{2}[2a + (17-1)d]$

$289 = \frac{17}{2}[2a + 16d]$

Multiply both sides by $\frac{2}{17}$:

$289 \times \frac{2}{17} = 2a + 16d$

$17 \times 2 = 2a + 16d$

$34 = 2a + 16d$

Divide the equation by 2:

Now we have a system of two linear equations:

$a + 3d = 7$ (1)

$a + 8d = 17$ (2)

Subtract equation (1) from equation (2):

$(a + 8d) - (a + 3d) = 17 - 7$

$a + 8d - a - 3d = 10$

$5d = 10$

Divide by 5:

$d = \frac{10}{5}$

$d = 2$

Substitute the value of $d = 2$ into equation (1):

$a + 3(2) = 7$

$a + 6 = 7$

$a = 7 - 6$

$a = 1$

So, the first term of the AP is $a=1$ and the common difference is $d=2$.

Now, we need to find the sum of the first $n$ terms, $S_n$. Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $a=1$ and $d=2$:

$S_n = \frac{n}{2}[2(1) + (n-1)2]$

$S_n = \frac{n}{2}[2 + 2n - 2]$

$S_n = \frac{n}{2}[2n]$

$S_n = n \times n$

$S_n = n^2$

The sum of the first $n$ terms of the AP is $n^2$.

To show that a sequence $a_1, a_2, \dots, a_n, \dots$ forms an Arithmetic Progression (AP), we need to show that the difference between consecutive terms, $a_{n+1} - a_n$, is a constant for all $n \geq 1$. This constant difference is the common difference ($d$) of the AP.

The sum of the first $k$ terms of an AP with first term $a$ and common difference $d$ is given by the formula $S_k = \frac{k}{2}[2a + (k-1)d]$.

---

(i) a_n = 3 + 4n

Given:

The $n$-th term is defined by $a_n = 3 + 4n$.

We need to find the sum of the first 15 terms.

To Show:

The sequence forms an AP.

To Find:

The sum of the first 15 terms ($S_{15}$).

Solution:

Let's find the $(n+1)$-th term, $a_{n+1}$, by replacing $n$ with $(n+1)$ in the formula for $a_n$:

$a_{n+1} = 3 + 4(n+1) = 3 + 4n + 4 = 7 + 4n$

Now, calculate the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (7 + 4n) - (3 + 4n)$

$a_{n+1} - a_n = 7 + 4n - 3 - 4n$

$a_{n+1} - a_n = 4$

Since the difference $a_{n+1} - a_n$ is a constant (4) and does not depend on $n$, the sequence $a_1, a_2, \dots, a_n, \dots$ forms an AP.

The common difference is $d = 4$.

To find the sum of the first 15 terms, we need the first term ($a_1$) and the common difference ($d$).

The first term is $a_1 = 3 + 4(1) = 3 + 4 = 7$. So, $a = 7$.

The common difference is $d = 4$.

The number of terms for the sum is $n = 15$.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{15} = \frac{15}{2}[2(7) + (15-1)4]$

$S_{15} = \frac{15}{2}[14 + (14)4]$

$S_{15} = \frac{15}{2}[14 + 56]$

$S_{15} = \frac{15}{2}[70]$

$S_{15} = 15 \times \frac{70}{2}$

$S_{15} = 15 \times 35$

$S_{15} = 525$

The sum of the first 15 terms is 525.

---

(ii) a_n = 9 – 5n

Given:

The $n$-th term is defined by $a_n = 9 - 5n$.

We need to find the sum of the first 15 terms.

To Show:

The sequence forms an AP.

To Find:

The sum of the first 15 terms ($S_{15}$).

Solution:

Let's find the $(n+1)$-th term, $a_{n+1}$, by replacing $n$ with $(n+1)$ in the formula for $a_n$:

$a_{n+1} = 9 - 5(n+1) = 9 - 5n - 5 = 4 - 5n$

Now, calculate the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (4 - 5n) - (9 - 5n)$

$a_{n+1} - a_n = 4 - 5n - 9 + 5n$

$a_{n+1} - a_n = -5$

Since the difference $a_{n+1} - a_n$ is a constant (-5) and does not depend on $n$, the sequence $a_1, a_2, \dots, a_n, \dots$ forms an AP.

The common difference is $d = -5$.

To find the sum of the first 15 terms, we need the first term ($a_1$) and the common difference ($d$).

The first term is $a_1 = 9 - 5(1) = 9 - 5 = 4$. So, $a = 4$.

The common difference is $d = -5$.

The number of terms for the sum is $n = 15$.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_{15} = \frac{15}{2}[2(4) + (15-1)(-5)]$

$S_{15} = \frac{15}{2}[8 + (14)(-5)]$

$S_{15} = \frac{15}{2}[8 - 70]$

$S_{15} = \frac{15}{2}[-62]$

$S_{15} = 15 \times \frac{-62}{2}$

$S_{15} = 15 \times (-31)$

$S_{15} = -465$

The sum of the first 15 terms is -465.

Given:

The sum of the first $n$ terms of an AP is given by the formula $S_n = 4n - n^2$.

---

To Find:

The first term ($a_1$), the sum of the first two terms ($S_2$), the second term ($a_2$), the 3rd term ($a_3$), the 10th term ($a_{10}$), and the $n$-th term ($a_n$).

---

Solution:

The sum of the first $n$ terms is given by $S_n = 4n - n^2$.

The first term ($a_1$) is equal to the sum of the first term ($S_1$).

Substitute $n=1$ into the formula for $S_n$:

$S_1 = 4(1) - (1)^2$

$S_1 = 4 - 1$

$S_1 = 3$

So, the first term is $a_1 = S_1 = 3$.

---

The sum of the first two terms ($S_2$) is found by substituting $n=2$ into the formula for $S_n$:

$S_2 = 4(2) - (2)^2$

$S_2 = 8 - 4$

$S_2 = 4$

---

The second term ($a_2$) can be found using the relationship that the second term is the sum of the first two terms minus the sum of the first term ($a_2 = S_2 - S_1$).

$a_2 = S_2 - S_1$

$a_2 = 4 - 3$

$a_2 = 1$

The second term is 1.

---

The common difference ($d$) of the AP can be found using the first two terms: $d = a_2 - a_1$.

$d = 1 - 3$

$d = -2$

The common difference is -2.

---

The 3rd term ($a_3$) can be found using the formula $a_n = a + (n-1)d$ or $a_3 = S_3 - S_2$. Let's use $a_n = a + (n-1)d$ with $a=a_1=3$ and $d=-2$:

$a_3 = a_1 + (3-1)d$

$a_3 = 3 + (2)(-2)$

$a_3 = 3 - 4$

$a_3 = -1$

The 3rd term is -1.

---

The 10th term ($a_{10}$) can be found using the formula $a_n = a + (n-1)d$ with $a=a_1=3$ and $d=-2$:

$a_{10} = a_1 + (10-1)d$

$a_{10} = 3 + (9)(-2)$

$a_{10} = 3 - 18$

$a_{10} = -15$

The 10th term is -15.

---

The $n$-th term ($a_n$) can be found using the relationship $a_n = S_n - S_{n-1}$ for $n > 1$.

We are given $S_n = 4n - n^2$.

To find $S_{n-1}$, substitute $(n-1)$ for $n$ in the formula for $S_n$:

$S_{n-1} = 4(n-1) - (n-1)^2$

$S_{n-1} = 4n - 4 - (n^2 - 2n + 1)$

$S_{n-1} = 4n - 4 - n^2 + 2n - 1$

$S_{n-1} = -n^2 + 6n - 5$

Now, calculate $a_n = S_n - S_{n-1}$ for $n > 1$:

$a_n = (4n - n^2) - (-n^2 + 6n - 5)$

$a_n = 4n - n^2 + n^2 - 6n + 5$

$a_n = (4n - 6n) + (-n^2 + n^2) + 5$

$a_n = -2n + 5$

Let's check if this formula works for $n=1$:

$a_1 = -2(1) + 5 = -2 + 5 = 3$. This matches our calculation of the first term.

Thus, the $n$-th term is given by $a_n = 5 - 2n$.

---

Summary of results:

First term ($a_1$) = 3

Sum of first two terms ($S_2$) = 4

Second term ($a_2$) = 1

3rd term ($a_3$) = -1

10th term ($a_{10}$) = -15

$n$-th term ($a_n$) = $5 - 2n$

Given:

We need to find the sum of the first 40 positive integers that are divisible by 6.

The sequence of positive integers divisible by 6 is 6, 12, 18, 24, . . .

---

To Find:

The sum of the first 40 terms of this sequence.

---

Solution:

The sequence of positive integers divisible by 6 forms an Arithmetic Progression (AP).

The first term is the smallest positive integer divisible by 6, which is $a = 6$.

The common difference is the constant increase, which is also 6, since we are considering multiples of 6. So, $d = 6$.

We need to find the sum of the first 40 terms, so the number of terms is $n = 40$.

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $n=40$, $a=6$, and $d=6$ into the formula:

$S_{40} = \frac{40}{2}[2(6) + (40-1)6]$

$S_{40} = 20[12 + (39)6]$

Calculate $39 \times 6$:

$39 \times 6 = (40 - 1) \times 6 = 240 - 6 = 234$

$S_{40} = 20[12 + 234]$

$S_{40} = 20[246]$

Calculate $20 \times 246$:

$20 \times 246 = 4920$

$S_{40} = 4920$

The sum of the first 40 positive integers divisible by 6 is 4920.

Given:

We need to find the sum of the first 15 positive integers that are multiples of 8.

The sequence of the first 15 multiples of 8 is 8, 16, 24, . . .

---

To Find:

The sum of the first 15 terms of this sequence.

---

Solution:

The sequence of the first 15 multiples of 8 forms an Arithmetic Progression (AP).

The first term is the first multiple of 8, which is $a = 8$.

The common difference is the difference between consecutive multiples of 8, which is $d = 16 - 8 = 8$.

We need to find the sum of the first 15 terms, so the number of terms is $n = 15$.

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $n=15$, $a=8$, and $d=8$ into the formula:

$S_{15} = \frac{15}{2}[2(8) + (15-1)8]$

$S_{15} = \frac{15}{2}[16 + (14)8]$

$S_{15} = \frac{15}{2}[16 + 112]$

$S_{15} = \frac{15}{2}[128]$

Simplify the expression:

$S_{15} = 15 \times \frac{128}{2}$

$S_{15} = 15 \times 64$

$S_{15} = 960$

The sum of the first 15 multiples of 8 is 960.

Given:

We need to find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50 are 1, 3, 5, . . ., 49.

---

To Find:

The sum of these odd numbers.

---

Solution:

The sequence of odd numbers between 0 and 50 is 1, 3, 5, . . ., 49.

This sequence forms an Arithmetic Progression (AP).

The first term is $a = 1$.

The common difference is $d = 3 - 1 = 2$.

The last term is $a_n = 49$.

First, we find the number of terms ($n$) in this AP using the formula $a_n = a + (n-1)d$.

$49 = 1 + (n-1)2$

Subtract 1 from both sides:

$49 - 1 = (n-1)2$

$48 = (n-1)2$

Divide both sides by 2:

$\frac{48}{2} = n-1$

$24 = n-1$

Add 1 to both sides:

$n = 24 + 1$

$n = 25$

There are 25 terms in the sequence.

Now, we find the sum of these 25 terms ($S_{25}$) using the formula for the sum of an AP when the first and last terms are known: $S_n = \frac{n}{2}(a + a_n)$.

Substitute the values $n=25$, $a=1$, and $a_n=49$ into the formula:

$S_{25} = \frac{25}{2}(1 + 49)$

$S_{25} = \frac{25}{2}(50)$

Simplify the expression:

$S_{25} = 25 \times \frac{50}{2}$

$S_{25} = 25 \times 25$

$S_{25} = 625$

The sum of the odd numbers between 0 and 50 is 625.

Given:

Penalty for the first day = $\textsf{₹} 200$.

Increase in penalty for each succeeding day = $\textsf{₹} 50$.

Number of days of delay = 30 days.

---

To Find:

The total amount of penalty the contractor has to pay for a delay of 30 days.

---

Solution:

The penalties for each day of delay form an Arithmetic Progression (AP).

The penalty for the first day is the first term, $a = \textsf{₹} 200$.

The increase in penalty each day is the common difference, $d = \textsf{₹} 50$.

The number of days of delay is the number of terms in the sum, $n = 30$.

The total penalty is the sum of the penalties for each of the 30 days, which is the sum of the first 30 terms of this AP ($S_{30}$).

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $n=30$, $a=200$, and $d=50$ into the formula:

$S_{30} = \frac{30}{2}[2(200) + (30-1)50]$

$S_{30} = 15[400 + (29)50]$

$S_{30} = 15[400 + 1450]$

$S_{30} = 15[1850]$

Now, calculate the product $15 \times 1850$:

$S_{30} = 27750$

The total penalty the contractor has to pay is $\textsf{₹} 27750$.

Given:

Total sum to be distributed as prizes = $\textsf{₹} 700$.

Number of cash prizes = 7.

Each prize is $\textsf{₹} 20$ less than its preceding prize.

---

To Find:

The value of each of the seven prizes.

---

Solution:

The values of the seven cash prizes form an Arithmetic Progression (AP).

Let the value of the first prize be $a$.

Since each succeeding prize is $\textsf{₹} 20$ less than the preceding one, the common difference is $d = -20$.

The number of prizes is $n = 7$.

The total sum of the prizes is the sum of the first 7 terms of this AP, $S_7 = \textsf{₹} 700$.

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the given values $S_7 = 700$, $n=7$, and $d=-20$ into the formula:

$700 = \frac{7}{2}[2a + (7-1)(-20)]$

$700 = \frac{7}{2}[2a + (6)(-20)]$

$700 = \frac{7}{2}[2a - 120]$

Multiply both sides by $\frac{2}{7}$:

$700 \times \frac{2}{7} = 2a - 120$

$\cancel{700}^{100} \times \frac{2}{\cancel{7}_{1}} = 2a - 120$

$100 \times 2 = 2a - 120$

$200 = 2a - 120$

Add 120 to both sides:

$200 + 120 = 2a$

$320 = 2a$

Divide by 2:

$a = \frac{320}{2}$

$a = 160$

The value of the first prize is $\textsf{₹} 160$.

Now, we can find the value of each of the other prizes by subtracting the common difference $\textsf{₹} 20$ successively.

The prizes are:

1st prize ($a_1$) = $a = \textsf{₹} 160$

2nd prize ($a_2$) = $a + d = 160 + (-20) = \textsf{₹} 140$

3rd prize ($a_3$) = $a + 2d = 160 + 2(-20) = 160 - 40 = \textsf{₹} 120$

4th prize ($a_4$) = $a + 3d = 160 + 3(-20) = 160 - 60 = \textsf{₹} 100$

5th prize ($a_5$) = $a + 4d = 160 + 4(-20) = 160 - 80 = \textsf{₹} 80$

6th prize ($a_6$) = $a + 5d = 160 + 5(-20) = 160 - 100 = \textsf{₹} 60$

7th prize ($a_7$) = $a + 6d = 160 + 6(-20) = 160 - 120 = \textsf{₹} 40$

The values of the seven cash prizes are $\textsf{₹} 160$, $\textsf{₹} 140$, $\textsf{₹} 120$, $\textsf{₹} 100$, $\textsf{₹} 80$, $\textsf{₹} 60$, and $\textsf{₹} 40$.

Given:

Classes in the school are from Class I to Class XII.

Number of sections in each class is 3.

Number of trees planted by each section of Class $k$ is equal to $k$.

---

To Find:

The total number of trees planted by the students of the school.

---

Solution:

For each class, there are 3 sections, and each section of Class $k$ plants $k$ trees.

So, the total number of trees planted by Class I = $1 \times 3 = 3$ trees.

The total number of trees planted by Class II = $2 \times 3 = 6$ trees.

The total number of trees planted by Class III = $3 \times 3 = 9$ trees.

... and so on, until Class XII.

The total number of trees planted by Class XII = $12 \times 3 = 36$ trees.

The number of trees planted by each class forms a sequence: 3, 6, 9, 12, . . ., 36.

This sequence is an Arithmetic Progression (AP) because the difference between consecutive terms is constant ($6 - 3 = 3$, $9 - 6 = 3$, etc.).

The first term of this AP is $a = 3$.

The common difference is $d = 3$.

The number of terms in this AP is the number of classes from I to XII, which is $n = 12$.

The last term of this AP is the number of trees planted by Class XII, $a_{12} = 36$.

The total number of trees planted is the sum of the first 12 terms of this AP ($S_{12}$).

We can use the formula for the sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}(a + a_n)$, where $a_n$ is the last term.

Substitute the values $n=12$, $a=3$, and $a_{12}=36$ into the formula:

$S_{12} = \frac{12}{2}(3 + 36)$

$S_{12} = 6(39)$

Calculate the product $6 \times 39$:

$S_{12} = 234$

The total number of trees planted by the students is 234.

Given:

A spiral is made up of 13 successive semicircles.

The radii of the semicircles are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .

The value of $\pi$ is $\frac{22}{7}$.

---

To Find:

The total length of the spiral made up of 13 consecutive semicircles.

---

Solution:

The length of a semicircle with radius $r$ is given by $\pi r$.

The radii of the successive semicircles are $r_1 = 0.5$, $r_2 = 1.0$, $r_3 = 1.5$, $r_4 = 2.0$, . . .

These radii form an Arithmetic Progression (AP) with first term $a_r = 0.5$ and common difference $d_r = 1.0 - 0.5 = 0.5$.

The length of the first semicircle is $l_1 = \pi r_1 = \pi (0.5) = 0.5\pi$.

The length of the second semicircle is $l_2 = \pi r_2 = \pi (1.0) = 1.0\pi$.

The length of the third semicircle is $l_3 = \pi r_3 = \pi (1.5) = 1.5\pi$.

The length of the fourth semicircle is $l_4 = \pi r_4 = \pi (2.0) = 2.0\pi$.

The lengths of the successive semicircles are $0.5\pi, 1.0\pi, 1.5\pi, 2.0\pi, \dots$

This sequence of lengths also forms an Arithmetic Progression (AP).

The first term of this AP of lengths is $a_l = 0.5\pi$.

The common difference of this AP is $d_l = 1.0\pi - 0.5\pi = 0.5\pi$.

The spiral is made up of 13 consecutive semicircles, so the number of terms in this AP of lengths is $n = 13$.

The total length of the spiral is the sum of the lengths of the first 13 semicircles, which is the sum of the first 13 terms of the AP of lengths ($S_{13}$).

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Substitute the values $n=13$, $a=0.5\pi$, and $d=0.5\pi$ into the formula:

$S_{13} = \frac{13}{2}[2(0.5\pi) + (13-1)(0.5\pi)]$

$S_{13} = \frac{13}{2}[1.0\pi + (12)(0.5\pi)]$

$S_{13} = \frac{13}{2}[\pi + 6\pi]$

$S_{13} = \frac{13}{2}[7\pi]$

$S_{13} = \frac{13 \times 7 \times \pi}{2}$

$S_{13} = \frac{91\pi}{2}$

Now, substitute the given value of $\pi = \frac{22}{7}$:

$S_{13} = \frac{91}{2} \times \frac{22}{7}$

Perform the multiplication, cancelling common factors:

$S_{13} = \frac{\cancel{91}^{13}}{\cancel{2}_{1}} \times \frac{\cancel{22}^{11}}{\cancel{7}_{1}}$

$S_{13} = 13 \times 11$

$S_{13} = 143$

The total length of the spiral is 143 cm.

Given:

Total number of logs = 200.

Number of logs in the bottom row (1st row) = 20.

Number of logs in the second row = 19.

Number of logs in the third row = 18.

The number of logs in each succeeding row is 1 less than the preceding row.

---

To Find:

The number of rows in which the 200 logs are placed.

The number of logs in the top row.

---

Solution:

The number of logs in each row forms an Arithmetic Progression (AP): 20, 19, 18, . . .

The first term of this AP is $a = 20$.

The common difference is $d = 19 - 20 = -1$.

The total number of logs is the sum of the terms in this AP. Let $n$ be the number of rows. The sum of the first $n$ terms is given as $S_n = 200$.

The formula for the sum of the first $n$ terms of an AP is:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $S_n = 200$, $a = 20$, and $d = -1$ into the formula:

$200 = \frac{n}{2}[2(20) + (n-1)(-1)]$

$200 = \frac{n}{2}[40 - (n-1)]$

$200 = \frac{n}{2}[40 - n + 1]$

$200 = \frac{n}{2}[41 - n]$

Multiply both sides by 2:

$2 \times 200 = n(41 - n)$

$400 = 41n - n^2$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

$n^2 - 41n + 400 = 0$

Now, we solve this quadratic equation for $n$ by factoring. We look for two numbers that multiply to 400 and add up to -41. These numbers are -16 and -25.

So, we can factor the quadratic equation as:

$(n - 16)(n - 25) = 0$

This gives two possible values for $n$:

$n - 16 = 0 \implies n = 16$

or

$n - 25 = 0 \implies n = 25$

Both 16 and 25 are positive integers, but we need to check if they make sense in the context of the problem.

Let's find the number of logs in the top row ($a_n$) for each value of $n$ using the formula $a_n = a + (n-1)d$.

Case 1: If $n = 16$ rows.

The number of logs in the 16th row is $a_{16}$:

$a_{16} = a + (16-1)d$

$a_{16} = 20 + (15)(-1)$

$a_{16} = 20 - 15 = 5$

Having 5 logs in the top row is physically possible.

Case 2: If $n = 25$ rows.

The number of logs in the 25th row is $a_{25}$:

$a_{25} = a + (25-1)d$

$a_{25} = 20 + (24)(-1)$

$a_{25} = 20 - 24 = -4$

Having -4 logs in the top row is not physically possible. The number of logs in a row must be a non-negative integer. As the number of logs per row decreases by 1, it would reach 0 and then theoretically become negative if we continued, but physically the stack would end when the number of logs becomes 1 or 0.

Therefore, the only valid number of rows is $n = 16$.

The number of logs in the top row is the number of logs in the 16th row, which is 5.

The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

Given:

Distance of the first potato from the bucket = $5$ m.

Distance between consecutive potatoes = $3$ m.

Number of potatoes = $10$.

---

To Find:

The total distance the competitor has to run.

---

Solution:

The competitor starts from the bucket, runs to the first potato, picks it up, and runs back to the bucket.

The distance covered to pick up the first potato and drop it in the bucket is twice the distance between the bucket and the first potato.

Distance run for the first potato = $2 \times 5$ m = $10$ m.

The competitor then runs from the bucket to the second potato, picks it up, and runs back to the bucket.

The distance of the second potato from the bucket is the distance of the first potato plus the distance between the first and second potato.

Distance of the second potato from the bucket = $5$ m $+ 3$ m $= 8$ m.

The distance covered to pick up the second potato and drop it in the bucket is twice the distance between the bucket and the second potato.

Distance run for the second potato = $2 \times 8$ m = $16$ m.

Similarly, the distance of the third potato from the bucket is the distance of the second potato plus the distance between the second and third potato.

Distance of the third potato from the bucket = $8$ m $+ 3$ m $= 11$ m.

The distance covered to pick up the third potato and drop it in the bucket is twice the distance between the bucket and the third potato.

Distance run for the third potato = $2 \times 11$ m = $22$ m.

The distances run by the competitor to pick up each potato and drop it into the bucket form a sequence:

$10, 16, 22, \dots$

This sequence is an Arithmetic Progression (AP) because the difference between consecutive terms is constant.

The first term of the AP is $a = 10$.

The common difference of the AP is $d = 16 - 10 = 6$.

There are $10$ potatoes, so there are $10$ terms in this AP ($n=10$), corresponding to the distance run for each potato.

The total distance run by the competitor is the sum of the first $10$ terms of this AP.

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substituting the values $n=10$, $a=10$, and $d=6$ into the formula:

$S_{10} = \frac{10}{2}[2(10) + (10-1)6]$

$S_{10} = 5[20 + (9)6]$

$S_{10} = 5[20 + 54]$

$S_{10} = 5[74]$

$S_{10} = 370$

---

Therefore, the total distance the competitor has to run is $370$ meters.

Given:

The given Arithmetic Progression (AP) is $121, 117, 113, \dots$

---

To Find:

The term number ($n$) for which the term ($a_n$) is the first negative term.

---

Solution:

From the given AP, the first term is $a = 121$.

The common difference $d$ is the difference between any term and its preceding term.

$d = 117 - 121 = -4$.

We want to find the first term that is negative. Let the $n$-th term be $a_n$. We need to find the smallest integer $n$ such that $a_n < 0$.

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

Substituting the values of $a$ and $d$, we get:

$a_n = 121 + (n-1)(-4)$

$a_n = 121 - 4(n-1)$

$a_n = 121 - 4n + 4$

$a_n = 125 - 4n$

Now, we set the condition for the term to be negative:

$a_n < 0$

$125 - 4n < 0$

To solve for $n$, we can rearrange the inequality:

$125 < 4n$

Divide both sides by 4:

$\frac{125}{4} < n$

Calculate the value of $\frac{125}{4}$:

$125 \div 4 = 31.25$

So, the inequality becomes:

$31.25 < n$

This means $n$ must be greater than $31.25$. Since $n$ represents the term number, it must be a positive integer ($n=1, 2, 3, \dots$).

The smallest integer value of $n$ that is greater than $31.25$ is $32$.

Thus, the 32nd term is the first term that is negative.

Let's verify the 31st and 32nd terms:

$a_{31} = 125 - 4(31) = 125 - 124 = 1$ (which is positive)

$a_{32} = 125 - 4(32) = 125 - 128 = -3$ (which is negative)

So, the 32nd term is indeed the first negative term.

---

The first negative term of the AP is the 32nd term.

Given:

Sum of the third and seventh terms of an AP is $6$.

Product of the third and seventh terms of the same AP is $8$.

---

To Find:

The sum of the first sixteen terms of the AP ($S_{16}$).

---

Solution:

Let the first term of the AP be $a$ and the common difference be $d$.

The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

The third term is $a_3 = a + (3-1)d = a + 2d$.

The seventh term is $a_7 = a + (7-1)d = a + 6d$.

According to the problem, the sum of the third and seventh terms is 6:

Substitute the expressions for $a_3$ and $a_7$:

$(a + 2d) + (a + 6d) = 6$

Combine like terms:

$2a + 8d = 6$

Divide the entire equation by 2:

This equation gives us a relationship between $a$ and $d$. We can express $a$ in terms of $d$:

$a = 3 - 4d$

Also, the product of the third and seventh terms is 8:

Substitute the expression for $a$ from equation (1) into this product equation:

$( (3 - 4d) + 2d ) ( (3 - 4d) + 6d ) = 8$

Simplify the terms inside the parentheses:

$(3 - 2d)(3 + 2d) = 8$

Using the difference of squares formula, $(x-y)(x+y) = x^2 - y^2$ where $x=3$ and $y=2d$:

$3^2 - (2d)^2 = 8$

$9 - 4d^2 = 8$

Rearranging the terms to solve for $d^2$:

$9 - 8 = 4d^2$

$1 = 4d^2$

Divide by 4:

$d^2 = \frac{1}{4}$

Taking the square root of both sides gives the possible values for $d$:

$d = \pm\sqrt{\frac{1}{4}}$

$d = \pm\frac{1}{2}$

We have two possible values for the common difference, $d = \frac{1}{2}$ or $d = -\frac{1}{2}$. We find the corresponding value of the first term $a$ for each case using equation (1): $a = 3 - 4d$.

---

Case 1: When $d = \frac{1}{2}$

Substitute $d = \frac{1}{2}$ into $a = 3 - 4d$:

$a = 3 - 4(\frac{1}{2})$

$a = 3 - 2$

$a = 1$

So, for this case, the first term is $a=1$ and the common difference is $d=\frac{1}{2}$. The AP is $1, 1.5, 2, 2.5, 3, 3.5, 4, \dots$

We need to find the sum of the first sixteen terms ($S_{16}$). The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

For $n=16$, $a=1$, and $d=\frac{1}{2}$:

$S_{16} = \frac{16}{2}[2(1) + (16-1)\frac{1}{2}]$

$S_{16} = 8[2 + 15 \times \frac{1}{2}]$

$S_{16} = 8[2 + \frac{15}{2}]$

To add the terms inside the brackets, find a common denominator:

$S_{16} = 8[\frac{4}{2} + \frac{15}{2}]$

$S_{16} = 8[\frac{4+15}{2}]$

$S_{16} = 8[\frac{19}{2}]$

Multiply the terms:

$S_{16} = \cancel{8}^4 \times \frac{19}{\cancel{2}^1}$

$S_{16} = 4 \times 19$

$S_{16} = 76$

The sum of the first sixteen terms is $76$ when $a=1$ and $d=\frac{1}{2}$.

---

Case 2: When $d = -\frac{1}{2}$

Substitute $d = -\frac{1}{2}$ into $a = 3 - 4d$:

$a = 3 - 4(-\frac{1}{2})$

$a = 3 + 2$

$a = 5$

So, for this case, the first term is $a=5$ and the common difference is $d=-\frac{1}{2}$. The AP is $5, 4.5, 4, 3.5, \dots$

We need to find the sum of the first sixteen terms ($S_{16}$). The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

For $n=16$, $a=5$, and $d=-\frac{1}{2}$:

$S_{16} = \frac{16}{2}[2(5) + (16-1)(-\frac{1}{2})]$

$S_{16} = 8[10 + 15 \times (-\frac{1}{2})]$

$S_{16} = 8[10 - \frac{15}{2}]$

To combine the terms inside the brackets, find a common denominator:

$S_{16} = 8[\frac{20}{2} - \frac{15}{2}]$

$S_{16} = 8[\frac{20-15}{2}]$

$S_{16} = 8[\frac{5}{2}]$

Multiply the terms:

$S_{16} = \cancel{8}^4 \times \frac{5}{\cancel{2}^1}$

$S_{16} = 4 \times 5$

$S_{16} = 20$

The sum of the first sixteen terms is $20$ when $a=5$ and $d=-\frac{1}{2}$.

---

Since both sets of parameters $(a=1, d=1/2)$ and $(a=5, d=-1/2)$ satisfy the given conditions regarding the third and seventh terms, there are two possible Arithmetic Progressions that fit the description.

The sum of the first sixteen terms of the AP can therefore be either $76$ or $20$.

Given:

Distance between consecutive rungs = $25$ cm.

Length of the bottom rung = $45$ cm.

Length of the top rung = $25$ cm.

Vertical distance between the top and bottom rungs = $2\frac{1}{2}$ m.

---

To Find:

The total length of wood required for the rungs.

---

Solution:

First, we need to find the total number of rungs. The vertical distance between the top and bottom rungs is given as $2\frac{1}{2}$ m. We convert this distance to centimeters to match the unit of the rung spacing:

$2\frac{1}{2}$ m = $2.5$ m = $2.5 \times 100$ cm = $250$ cm.

The rungs are $25$ cm apart. The number of spaces between the rungs is the total distance between the top and bottom rungs divided by the spacing between rungs.

Number of spaces between rungs = $\frac{\text{Total distance}}{\text{Spacing between rungs}} = \frac{250}{25} = 10$.

Since there is one more rung than the number of spaces between them (like fence posts and the gaps between them), the total number of rungs is the number of spaces plus one.

Number of rungs = Number of spaces + 1 = $10 + 1 = 11$.

The lengths of the rungs decrease uniformly from the bottom to the top. This means the lengths form an Arithmetic Progression (AP).

The first term of this AP is the length of the bottom rung, $a = 45$ cm.

The last term of this AP is the length of the top rung, $l = 25$ cm.

The number of terms in this AP is the number of rungs, $n = 11$.

The total length of wood required for the rungs is the sum of the lengths of all the rungs. This is the sum of the AP.

The formula for the sum of an AP when the first term ($a$), the last term ($l$), and the number of terms ($n$) are known is:

$S_n = \frac{n}{2}(a + l)$

Substitute the values $n=11$, $a=45$, and $l=25$ into the formula:

$S_{11} = \frac{11}{2}(45 + 25)$

$S_{11} = \frac{11}{2}(70)$

$S_{11} = 11 \times \frac{70}{2}$

$S_{11} = 11 \times 35$

Calculate $11 \times 35$:

$11 \times 35 = 385$

$S_{11} = 385$

The total length of wood required is $385$ cm.

We can also express this in meters:

$385$ cm $= \frac{385}{100}$ m $= 3.85$ m.

---

The total length of the wood required for the rungs is $385$ cm or $3.85$ m.

Given:

The houses are numbered consecutively from $1$ to $49$.

The numbers of the houses form an Arithmetic Progression (AP) with first term $a=1$ and common difference $d=1$. The terms are $1, 2, 3, \dots, 49$.

The total number of houses is $N=49$.

---

To Find:

A value of $x$ (where $1 < x \le 49$) such that the sum of the numbers of the houses preceding house $x$ is equal to the sum of the numbers of the houses following house $x$. Specifically, we need to find this value of $x$.

---

Solution:

Let $S_n$ denote the sum of the numbers of the first $n$ houses. Since the house numbers are $1, 2, \dots, n$, this is the sum of the first $n$ natural numbers.

The formula for the sum of the first $n$ natural numbers is $S_n = \frac{n(n+1)}{2}$.

The houses preceding the house numbered $x$ are $1, 2, \dots, x-1$. The sum of the numbers of these houses is $S_{x-1}$.

$S_{x-1} = \frac{(x-1)((x-1)+1)}{2} = \frac{(x-1)x}{2}$.

The houses following the house numbered $x$ are $x+1, x+2, \dots, 49$. The sum of the numbers of these houses can be found by taking the total sum of all house numbers from $1$ to $49$ ($S_{49}$) and subtracting the sum of house numbers from $1$ to $x$ ($S_x$).

Sum of houses following house $x = S_{49} - S_x$.

$S_{49} = \frac{49(49+1)}{2} = \frac{49 \times 50}{2}$.

$S_x = \frac{x(x+1)}{2}$.

According to the problem statement, the sum of the numbers of the houses preceding $x$ is equal to the sum of the numbers of the houses following $x$.

Using our notation, this translates to:

Substitute the formulas for the sums into equation (1):

$\frac{(x-1)x}{2} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2}$

To eliminate the denominators, multiply the entire equation by 2:

$(x-1)x = 49 \times 50 - x(x+1)$

Expand the terms:

$x^2 - x = 2450 - (x^2 + x)$

$x^2 - x = 2450 - x^2 - x$

Add $x$ to both sides:

$x^2 = 2450 - x^2$

Add $x^2$ to both sides:

$x^2 + x^2 = 2450$

$2x^2 = 2450$

Divide both sides by 2:

$x^2 = \frac{2450}{2}$

$x^2 = 1225$

Take the square root of both sides:

$x = \pm\sqrt{1225}$

Calculate the square root of 1225:

$\sqrt{1225} = 35$

So, $x = \pm 35$.

Since $x$ represents a house number, it must be a positive integer. Therefore, we take the positive value.

$x = 35$

We check if $x=35$ is within the range of house numbers $1$ to $49$. $1 \le 35 \le 49$, which is true.

Also, for the sum of preceding houses $S_{x-1}$ to be meaningful, we need $x-1 \ge 1$, which means $x \ge 2$. For the sum of following houses to be meaningful, we need $x+1 \le 49$, which means $x \le 48$. Our value $x=35$ satisfies $2 \le 35 \le 48$.

Thus, there exists a value of $x$ that satisfies the condition, and that value is 35.

To verify, let's calculate the sums:

Sum of houses preceding house 35: $S_{34} = \frac{34(34+1)}{2} = \frac{34 \times 35}{2} = 17 \times 35 = 595$.

Sum of houses from 1 to 49: $S_{49} = \frac{49(49+1)}{2} = \frac{49 \times 50}{2} = 49 \times 25 = 1225$.

Sum of houses from 1 to 35: $S_{35} = \frac{35(35+1)}{2} = \frac{35 \times 36}{2} = 35 \times 18 = 630$.

Sum of houses following house 35 = $S_{49} - S_{35} = 1225 - 630 = 595$.

Since $S_{34} = 595$ and $S_{49} - S_{35} = 595$, the condition is satisfied for $x=35$.

---

The value of $x$ is $35$.

Given:

Number of steps = 15.

Length of each step = 50 m.

Rise (height) of each step = $\frac{1}{4}$ m.

Tread (depth) of each step = $\frac{1}{2}$ m.

---

To Find:

The total volume of concrete required to build the terrace.

---

Solution:

Based on the figure and the description, the steps are built such that the concrete forming each step has a constant height (rise) and length, but the width (tread) of the concrete *layer* for each subsequent step increases. The first step has a tread of $\frac{1}{2}$ m. The second step adds another layer of concrete with a tread of $\frac{1}{2}$ m on top, making the concrete for the second step have a total effective tread of $1$ m. The third step adds a layer on top of that, making the concrete for the third step have a total effective tread of $\frac{3}{2}$ m, and so on.

The volume of concrete for each step can be considered as a rectangular prism with dimensions: Length $\times$ Tread (effective width) $\times$ Rise.

• Volume of concrete for the 1st step: Length = 50m, Tread = $\frac{1}{2}$m, Rise = $\frac{1}{4}$m.

Volume of 1st step $= 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{50}{8} = \frac{25}{4}$ m$^3$.

• Volume of concrete for the 2nd step: Length = 50m, Tread = $1$m, Rise = $\frac{1}{4}$m.

Volume of 2nd step $= 50 \times 1 \times \frac{1}{4} = \frac{50}{4} = \frac{25}{2}$ m$^3$.

• Volume of concrete for the 3rd step: Length = 50m, Tread = $\frac{3}{2}$m, Rise = $\frac{1}{4}$m.

Volume of 3rd step $= 50 \times \frac{3}{2} \times \frac{1}{4} = \frac{150}{8} = \frac{75}{4}$ m$^3$.

The volumes of concrete required for the 15 steps form a sequence:

$\frac{25}{4}, \frac{25}{2}, \frac{75}{4}, \dots$

Let's check if this sequence is an Arithmetic Progression (AP) by finding the difference between consecutive terms:

Difference between 2nd and 1st term $= \frac{25}{2} - \frac{25}{4} = \frac{50}{4} - \frac{25}{4} = \frac{25}{4}$

Difference between 3rd and 2nd term $= \frac{75}{4} - \frac{25}{2} = \frac{75}{4} - \frac{50}{4} = \frac{25}{4}$

Since the difference between consecutive terms is constant ($\frac{25}{4}$), the sequence of volumes forms an AP.

The first term of this AP is $a = \frac{25}{4}$ m$^3$.

The common difference is $d = \frac{25}{4}$ m$^3$.

There are 15 steps, so the number of terms in this AP is $n = 15$.

The total volume of concrete required is the sum of the volumes of the 15 steps, which is the sum of the first 15 terms of this AP. We use the formula for the sum of an AP:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values $n=15$, $a=\frac{25}{4}$, and $d=\frac{25}{4}$:

$S_{15} = \frac{15}{2}[2(\frac{25}{4}) + (15-1)(\frac{25}{4})]$

$S_{15} = \frac{15}{2}[\frac{50}{4} + 14(\frac{25}{4})]$

$S_{15} = \frac{15}{2}[\frac{50}{4} + \frac{14 \times 25}{4}]$

$S_{15} = \frac{15}{2}[\frac{50}{4} + \frac{350}{4}]$

$S_{15} = \frac{15}{2}[\frac{50 + 350}{4}]$

$S_{15} = \frac{15}{2}[\frac{400}{4}]$

$S_{15} = \frac{15}{2}[100]$

$S_{15} = 15 \times \frac{100}{2}$

$S_{15} = 15 \times 50$

$S_{15} = 750$

The total volume of concrete required is 750 m$^3$.

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Answer:

The total volume of concrete required to build the terrace is 750 m$^3$.