Chapter 5 Arithmetic Progressions

Given:

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

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To Find:

The first term ($a$) and the common difference ($d$).

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Solution:

The first term ($a$) of an AP is the first number in the sequence.

In this AP, the first term is $a = \frac{3}{2}$.

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

We can find the common difference by subtracting the first term from the second term:

$d = a_2 - a_1 = \frac{1}{2} - \frac{3}{2}$

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

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

$d = -1$

We can verify this by subtracting the second term from the third term:

$d = a_3 - a_2 = -\frac{1}{2} - \frac{1}{2}$

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

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

$d = -1$

The common difference is consistent.

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Final Answer:

For the given AP, the first term $a = \mathbf{\frac{3}{2}}$ and the common difference $d = \mathbf{-1}$.

General Concept: Arithmetic Progression (AP)

A list of numbers $a_1, a_2, a_3, \dots$ is an Arithmetic Progression (AP) if the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

$d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots$

If a list forms an AP, the next term ($a_{n+1}$) is found by adding the common difference ($d$) to the last known term ($a_n$).

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

Given list: 4, 10, 16, 22, ...

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

Calculate the differences 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), the list of numbers forms an AP.

The common difference $d = 6$.

The last given term is $a_4 = 22$.

The next term ($a_5$) = $a_4 + d = 22 + 6 = 28$.

The term after that ($a_6$) = $a_5 + d = 28 + 6 = 34$.

Conclusion: The list forms an AP with a common difference of 6. The next two terms are 28 and 34.

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Solution (ii): 1, -1, -3, -5, ...

Given list: 1, -1, -3, -5, ...

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

Calculate the differences 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), the list of numbers forms an AP.

The common difference $d = -2$.

The last given term is $a_4 = -5$.

The next term ($a_5$) = $a_4 + d = -5 + (-2) = -7$.

The term after that ($a_6$) = $a_5 + d = -7 + (-2) = -9$.

Conclusion: The list forms an AP with a common difference of -2. The next two terms are -7 and -9.

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Solution (iii): -2, 2, -2, 2, -2, ...

Given list: -2, 2, -2, 2, -2, ...

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

Calculate the differences between consecutive terms:

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

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

Since the difference between the second and first term (4) is not equal to the difference between the third and second term (-4), the differences are not constant.

Conclusion: The list of numbers does not form an AP.

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

Given list: 1, 1, 1, 2, 2, 2, 3, 3, 3, ...

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

Calculate the differences 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 the fourth and third term (1) is not equal to the difference between the third and second term (0), the differences are not constant.

Conclusion: The list of numbers does not form an AP.

General Concept: Arithmetic Progression (AP)

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number (the common difference) to the preceding term, except the first term.

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Solution (i): Taxi Fare

Let $a_n$ be the taxi fare after $n$ km.

Fare for the 1st km ($a_1$) = $\textsf{₹} 15$.

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

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

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

The list of fares after each km is: 15, 23, 31, 39, ...

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 ($d = 8$), the list of numbers involved makes an arithmetic progression.

Why: Each subsequent term (fare after the next km) is obtained by adding a fixed amount (₹ 8) to the previous term.

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Solution (ii): Air in Cylinder

Let the initial amount of air present in the cylinder be $V$ units.

Amount of air after 1st removal ($a_1$) = $V - \frac{1}{4}V = \frac{3}{4}V$.

Amount of air after 2nd removal ($a_2$) = $\frac{3}{4}V - \frac{1}{4}\left(\frac{3}{4}V\right) \ $$ = \frac{3}{4}V - \frac{3}{16}V = \left(\frac{12-3}{16}\right)V = \frac{9}{16}V$.

Amount of air after 3rd removal ($a_3$) = $\frac{9}{16}V - \frac{1}{4}\left(\frac{9}{16}V\right) \ $$ = \frac{9}{16}V - \frac{9}{64}V = \left(\frac{36-9}{64}\right)V = \frac{27}{64}V$.

The list of amounts of air remaining is: $\frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, \dots$ (assuming $a_1$ is after the first removal).

Check the difference between consecutive terms:

$a_2 - a_1 = \frac{9}{16}V - \frac{3}{4}V = \frac{9V - 12V}{16} = -\frac{3}{16}V$.

$a_3 - a_2 = \frac{27}{64}V - \frac{9}{16}V = \frac{27V - 36V}{64} = -\frac{9}{64}V$.

Since the difference between consecutive terms ($-\frac{3}{16}V$ and $-\frac{9}{64}V$) is not constant, the list of numbers involved does not make an arithmetic progression.

Why: Each subsequent term is obtained by multiplying the previous term by a fixed factor ($\frac{3}{4}$), not by adding a fixed amount.

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Solution (iii): Cost of Digging

Let $a_n$ be the cost of digging for the $n$-th metre.

Cost for the 1st metre ($a_1$) = $\textsf{₹} 150$.

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

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

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

The list of costs for digging each subsequent metre is: 150, 200, 250, 300, ...

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 ($d = 50$), the list of costs for each metre of digging makes an arithmetic progression.

Why: The cost for digging each subsequent metre increases by a fixed amount (₹ 50).

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Solution (iv): Compound Interest

Let $A_n$ be the amount of money in the account after $n$ years.

Principal ($P$) = $\textsf{₹} 10000$.

Rate ($r$) = 8% per annum.

Amount after 1 year ($A_1$) = $P\left(1 + \frac{r}{100}\right)^1 = 10000\left(1 + \frac{8}{100}\right)^1 \ $$ = 10000(1.08)^1 = \textsf{₹} 10800$.

Amount after 2 years ($A_2$) = $P\left(1 + \frac{r}{100}\right)^2 = 10000\left(1 + \frac{8}{100}\right)^2 \ $$ = 10000(1.08)^2 = 10000(1.1664) = \textsf{₹} 11664$.

Amount after 3 years ($A_3$) = $P\left(1 + \frac{r}{100}\right)^3 = 10000\left(1 + \frac{8}{100}\right)^3 \ $$ = 10000(1.08)^3 = 10000(1.259712) = \textsf{₹} 12597.12$.

The list of amounts after each year is: 10800, 11664, 12597.12, ...

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 (864 and 933.12) is not constant, the list of numbers involved does not make an arithmetic progression.

Why: In compound interest, the interest earned each year is added to the principal, and the next year's interest is calculated on the new, larger principal. Thus, the increase in the amount is not constant.

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 and $d$ is the common difference.

We need to find the first four terms, which are $a_1$, $a_2$, $a_3$, and $a_4$.

$a_1 = a + (1-1)d = a$

$a_2 = a + (2-1)d = a + d$

$a_3 = a + (3-1)d = a + 2d$

$a_4 = a + (4-1)d = a + 3d$

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(i) Given $a = 10$, $d = 10$.

First term, $a_1 = a = 10$.

Second term, $a_2 = a + d = 10 + 10 = 20$.

Third term, $a_3 = a + 2d = 10 + 2(10) = 10 + 20 = 30$.

Fourth term, $a_4 = a + 3d = 10 + 3(10) = 10 + 30 = 40$.

The first four terms of the AP are: $10, 20, 30, 40$.

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(ii) Given $a = -2$, $d = 0$.

First term, $a_1 = a = -2$.

Second term, $a_2 = a + d = -2 + 0 = -2$.

Third term, $a_3 = a + 2d = -2 + 2(0) = -2 + 0 = -2$.

Fourth term, $a_4 = a + 3d = -2 + 3(0) = -2 + 0 = -2$.

The first four terms of the AP are: $-2, -2, -2, -2$.

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(iii) Given $a = 4$, $d = -3$.

First term, $a_1 = a = 4$.

Second term, $a_2 = a + d = 4 + (-3) = 4 - 3 = 1$.

Third term, $a_3 = a + 2d = 4 + 2(-3) = 4 - 6 = -2$.

Fourth term, $a_4 = a + 3d = 4 + 3(-3) = 4 - 9 = -5$.

The first four terms of the AP are: $4, 1, -2, -5$.

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(iv) Given $a = -1$, $d = \frac{1}{2}$.

First term, $a_1 = a = -1$.

Second term, $a_2 = a + d = -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2}$.

Third term, $a_3 = a + 2d = -1 + 2(\frac{1}{2}) = -1 + 1 = 0$.

Fourth term, $a_4 = a + 3d = -1 + 3(\frac{1}{2}) = -1 + \frac{3}{2} = -\frac{2}{2} + \frac{3}{2} = \frac{1}{2}$.

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

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(v) Given $a = -1.25$, $d = -0.25$.

First term, $a_1 = a = -1.25$.

Second term, $a_2 = a + d = -1.25 + (-0.25) = -1.25 - 0.25 = -1.50$.

Third term, $a_3 = a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75$.

Fourth term, $a_4 = a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00$.

The first four terms of the AP are: $-1.25, -1.50, -1.75, -2.00$.

In an Arithmetic Progression (AP), the first term is the first number in the sequence, and the common difference is the constant difference between any term and its preceding term. The common difference $d$ can be found by subtracting any term from its subsequent term, i.e., $d = a_{n+1} - a_n$.

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(i) Given AP: $3, 1, -1, -3, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = 3$.

The common difference $d$ is the difference between consecutive terms.

$d = 1 - 3 = -2$.

Alternatively, $d = -1 - 1 = -2$, or $d = -3 - (-1) = -3 + 1 = -2$.

The first term is $3$ and the common difference is $-2$.

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(ii) Given AP: $-5, -1, 3, 7, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = -5$.

The common difference $d$ is the difference between consecutive terms.

$d = -1 - (-5) = -1 + 5 = 4$.

Alternatively, $d = 3 - (-1) = 3 + 1 = 4$, or $d = 7 - 3 = 4$.

The first term is $-5$ and the common difference is $4$.

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(iii) Given AP: $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$

The first term $a$ is the first number in the sequence.

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

The common difference $d$ is the difference between consecutive terms.

$d = \frac{5}{3} - \frac{1}{3} = \frac{5-1}{3} = \frac{4}{3}$.

Alternatively, $d = \frac{9}{3} - \frac{5}{3} = \frac{9-5}{3} = \frac{4}{3}$, or $d = \frac{13}{3} - \frac{9}{3} = \frac{13-9}{3} = \frac{4}{3}$.

The first term is $\frac{1}{3}$ and the common difference is $\frac{4}{3}$.

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(iv) Given AP: $0.6, 1.7, 2.8, 3.9, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = 0.6$.

The common difference $d$ is the difference between consecutive terms.

$d = 1.7 - 0.6 = 1.1$.

Alternatively, $d = 2.8 - 1.7 = 1.1$, or $d = 3.9 - 2.8 = 1.1$.

The first term is $0.6$ and the common difference is $1.1$.

A sequence is an Arithmetic Progression (AP) if the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$. To check if a sequence is an AP, we calculate the difference between successive terms ($a_{n+1} - a_n$) for consecutive pairs.

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(i) Given sequence: $2, 4, 8, 16, \dots$

Difference between 2nd and 1st term: $4 - 2 = 2$

Difference between 3rd and 2nd term: $8 - 4 = 4$

Since the differences are not equal ($2 \neq 4$), the sequence is not an AP.

Conclusion: Not an AP.

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(ii) Given sequence: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

Difference between 2nd and 1st term: $\frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$

Difference between 3rd and 2nd term: $3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}$

Difference between 4th and 3rd term: $\frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \frac{1}{2}$.

The first four terms are $a_1 = 2, a_2 = \frac{5}{2}, a_3 = 3, a_4 = \frac{7}{2}$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = \frac{1}{2}$. Next three terms: $4, \frac{9}{2}, 5$.

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(iii) Given sequence: $-1.2, -3.2, -5.2, -7.2, \dots$

Difference between 2nd and 1st term: $-3.2 - (-1.2) = -3.2 + 1.2 = -2.0$

Difference between 3rd and 2nd term: $-5.2 - (-3.2) = -5.2 + 3.2 = -2.0$

Difference between 4th and 3rd term: $-7.2 - (-5.2) = -7.2 + 5.2 = -2.0$

Since the difference is constant, the sequence is an AP with common difference $d = -2.0$.

The first four terms are $a_1 = -1.2, a_2 = -3.2, a_3 = -5.2, a_4 = -7.2$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = -2.0$. Next three terms: $-9.2, -11.2, -13.2$.

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(iv) Given sequence: $-10, -6, -2, 2, \dots$

Difference between 2nd and 1st term: $-6 - (-10) = -6 + 10 = 4$

Difference between 3rd and 2nd term: $-2 - (-6) = -2 + 6 = 4$

Difference between 4th and 3rd term: $2 - (-2) = 2 + 2 = 4$

Since the difference is constant, the sequence is an AP with common difference $d = 4$.

The first four terms are $a_1 = -10, a_2 = -6, a_3 = -2, a_4 = 2$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = 4$. Next three terms: $6, 10, 14$.

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(v) Given sequence: $3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots$

Difference between 2nd and 1st term: $(3 + \sqrt{2}) - 3 = \sqrt{2}$

Difference between 3rd and 2nd term: $(3 + 2\sqrt{2}) - (3 + \sqrt{2}) = 3 + 2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2}$

Difference between 4th and 3rd term: $(3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = 3 + 3\sqrt{2} - 3 - 2\sqrt{2} = \sqrt{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \sqrt{2}$.

The first four terms are $a_1 = 3, a_2 = 3+\sqrt{2}, a_3 = 3+2\sqrt{2}, a_4 = 3+3\sqrt{2}$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = \sqrt{2}$. Next three terms: $3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}$.

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(vi) Given sequence: $0.2, 0.22, 0.222, 0.2222, \dots$

Difference between 2nd and 1st term: $0.22 - 0.2 = 0.02$

Difference between 3rd and 2nd term: $0.222 - 0.22 = 0.002$

Since the differences are not equal ($0.02 \neq 0.002$), the sequence is not an AP.

Conclusion: Not an AP.

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(vii) Given sequence: $0, -4, -8, -12, \dots$

Difference between 2nd and 1st term: $-4 - 0 = -4$

Difference between 3rd and 2nd term: $-8 - (-4) = -8 + 4 = -4$

Difference between 4th and 3rd term: $-12 - (-8) = -12 + 8 = -4$

Since the difference is constant, the sequence is an AP with common difference $d = -4$.

The first four terms are $a_1 = 0, a_2 = -4, a_3 = -8, a_4 = -12$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = -4$. Next three terms: $-16, -20, -24$.

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(viii) Given sequence: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$

Difference between 2nd and 1st term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Difference between 3rd and 2nd term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Difference between 4th and 3rd term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Since the difference is constant, the sequence is an AP with common difference $d = 0$.

The first four terms are $a_1 = -\frac{1}{2}, a_2 = -\frac{1}{2}, a_3 = -\frac{1}{2}, a_4 = -\frac{1}{2}$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = 0$. Next three terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.

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(ix) Given sequence: $1, 3, 9, 27, \dots$

Difference between 2nd and 1st term: $3 - 1 = 2$

Difference between 3rd and 2nd term: $9 - 3 = 6$

Since the differences are not equal ($2 \neq 6$), the sequence is not an AP.

Conclusion: Not an AP.

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(x) Given sequence: $a, 2a, 3a, 4a, \dots$

Difference between 2nd and 1st term: $2a - a = a$

Difference between 3rd and 2nd term: $3a - 2a = a$

Difference between 4th and 3rd term: $4a - 3a = a$

Since the difference is constant (assuming $a$ is a constant value), the sequence is an AP with common difference $d = a$.

The first four terms are $a_1 = a, a_2 = 2a, a_3 = 3a, a_4 = 4a$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = a$. Next three terms: $5a, 6a, 7a$.

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(xi) Given sequence: $a, a^2, a^3, a^4, \dots$

Difference between 2nd and 1st term: $a^2 - a = a(a-1)$

Difference between 3rd and 2nd term: $a^3 - a^2 = a^2(a-1)$

For this to be an AP, $a(a-1)$ must be equal to $a^2(a-1)$.

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

$a^2(a-1) - a(a-1) = 0$

$a(a-1)(a - 1) = 0$

$a(a-1)^2 = 0$

This is only true if $a=0$ or $a=1$. For any other value of $a$, the difference is not constant.

Conclusion: Not an AP (unless $a=0$ or $a=1$, in which case it is an AP with $d=0$; however, as a general sequence based on $a$, it is not). Let's assume it's not an AP for arbitrary $a$.

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(xii) Given sequence: $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$

Rewrite the terms in simplest radical form:

$\sqrt{2}$

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

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

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

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

Difference between 2nd and 1st term: $2\sqrt{2} - \sqrt{2} = \sqrt{2}$

Difference between 3rd and 2nd term: $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

Difference between 4th and 3rd term: $4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \sqrt{2}$.

The first four terms are $a_1 = \sqrt{2}, a_2 = 2\sqrt{2}, a_3 = 3\sqrt{2}, a_4 = 4\sqrt{2}$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = \sqrt{2}$. Next three terms: $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

---

(xiii) Given sequence: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$

Rewrite the terms: $\sqrt{3}, \sqrt{6}, 3, \sqrt{12} = 2\sqrt{3}$

Difference between 2nd and 1st term: $\sqrt{6} - \sqrt{3}$

Difference between 3rd and 2nd term: $3 - \sqrt{6}$

To check if $\sqrt{6} - \sqrt{3} = 3 - \sqrt{6}$, we add $\sqrt{6}$ and $\sqrt{3}$ to both sides:

$2\sqrt{6} = 3 + \sqrt{3}$

Squaring both sides: $(2\sqrt{6})^2 = (3 + \sqrt{3})^2$

$4 \times 6 = 3^2 + 2(3)\sqrt{3} + (\sqrt{3})^2$

$24 = 9 + 6\sqrt{3} + 3$

$24 = 12 + 6\sqrt{3}$

$12 = 6\sqrt{3}$

$2 = \sqrt{3}$

This is false ($2^2 = 4$, $(\sqrt{3})^2 = 3$). Therefore, the differences are not equal.

Conclusion: Not an AP.

---

(xiv) Given sequence: $1^2, 3^2, 5^2, 7^2, \dots$

Rewrite the terms: $1, 9, 25, 49, \dots$

Difference between 2nd and 1st term: $9 - 1 = 8$

Difference between 3rd and 2nd term: $25 - 9 = 16$

Since the differences are not equal ($8 \neq 16$), the sequence is not an AP.

Conclusion: Not an AP.

---

(xv) Given sequence: $1^2, 5^2, 7^2, 73, \dots$

Rewrite the terms: $1, 25, 49, 73, \dots$

Difference between 2nd and 1st term: $25 - 1 = 24$

Difference between 3rd and 2nd term: $49 - 25 = 24$

Difference between 4th and 3rd term: $73 - 49 = 24$

Since the difference is constant, the sequence is an AP with common difference $d = 24$.

The first four terms are $a_1 = 1, a_2 = 25, a_3 = 49, a_4 = 73$.

The next three terms are:

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

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

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

Conclusion: It is an AP. Common difference $d = 24$. Next three terms: $97, 121, 145$.

The given Arithmetic Progression (AP) is: $2, 7, 12, \dots$

---

Here, the first term is $a = 2$.

---

The common difference $d$ is the difference between consecutive terms.

$d = 7 - 2 = 5$.

We can verify this with the next pair of terms: $12 - 7 = 5$.

---

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

---

We need to find the 10th term of this AP, so we need to find $a_{10}$. Here, $n = 10$.

Substitute $a=2$, $d=5$, and $n=10$ into the formula (i):

$a_{10} = 2 + (10-1) \times 5$

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

$a_{10} = 2 + 45$

$a_{10} = 47$

---

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

The given Arithmetic Progression (AP) is: $21, 18, 15, \dots$

---

Here, the first term is $a = 21$.

---

The common difference $d$ is the difference between consecutive terms.

$d = 18 - 21 = -3$.

We can verify this with the next pair of terms: $15 - 18 = -3$. So, $d = -3$.

---

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

---

Part 1: Which term is -81?

We want to find the value of $n$ for which $a_n = -81$.

Substitute $a = 21$, $d = -3$, and $a_n = -81$ into formula (i):

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

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

---

Part 2: Is any term 0?

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

Substitute $a = 21$, $d = -3$, and $a_m = 0$ into formula (i) (using $m$ instead of $n$):

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

Subtract 21 from both sides:

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

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

Divide both sides by -3:

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

$7 = m-1$

Add 1 to both sides:

$m = 7 + 1$

$m = 8$

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

Reason: A number is a term in an AP if and only if the index $n$ calculated using the formula $a_n = a + (n-1)d$ is a positive integer. In both cases ($a_n = -81$ and $a_n = 0$), we obtained positive integer values for $n$ (35 and 8 respectively), which means both $-81$ and $0$ are terms in the given AP.

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

---

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

---

We are given that the 3rd term of the AP is 5. Using formula (1) with $n=3$:

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

$a_3 = a + 2d$

Since $a_3 = 5$, we have the equation:

---

We are also given that the 7th term of the AP is 9. Using formula (1) with $n=7$:

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

$a_7 = a + 6d$

Since $a_7 = 9$, we have the equation:

---

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

$a + 2d = 5$

$a + 6d = 9$

To solve for $d$, we can subtract equation (i) from equation (ii):

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

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

$4d = 4$

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

---

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

$a + 2(1) = 5$

$a + 2 = 5$

$a = 5 - 2$

---

The Arithmetic Progression is determined by its first term ($a$) and its common difference ($d$).

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

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

Substituting the values of $a$ and $d$, the AP is:

$3, 3+1, 3+2(1), 3+3(1), \dots$

which is:

$3, 4, 5, 6, \dots$

---

The AP is $\mathbf{3, 4, 5, 6, \dots}$.

The given list of numbers is: $5, 11, 17, 23, \dots$

---

First, we check if this list forms an Arithmetic Progression (AP) by finding the difference between consecutive terms.

Difference between 2nd and 1st term: $11 - 5 = 6$

Difference between 3rd and 2nd term: $17 - 11 = 6$

Difference between 4th and 3rd term: $23 - 17 = 6$

Since the difference is constant, the list of numbers is an AP.

---

The first term of this AP is $a = 5$.

The common difference is $d = 6$.

---

To check if 301 is a term of this AP, we need to determine if there exists a positive integer $n$ such that the $n$-th term, $a_n$, is equal to 301.

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

---

We set $a_n = 301$ and substitute the values of $a$ and $d$ into the formula (1):

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

Now, we solve this equation for $n$:

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

$296 = (n-1)6$

Divide both sides by 6:

Simplify the fraction on the left side:

$\frac{\cancel{296}^{148}}{\cancel{6}_{3}} = \frac{148}{3}$

So, equation (i) becomes:

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

Add 1 to both sides to find $n$:

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

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

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

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

---

For 301 to be a term of the AP, the value of $n$ must be a positive integer.

The value we obtained for $n$ is $\frac{151}{3}$.

Since $\frac{151}{3}$ is not an integer (151 divided by 3 leaves a remainder), 301 is not a term of the given AP.

---

Conclusion: 301 is not a term of the list of numbers $5, 11, 17, 23, \dots$.

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

The two-digit numbers start from 10 and end at 99.

---

The first two-digit number divisible by 3 is 12.

The next two-digit number divisible by 3 is 15.

The numbers divisible by 3 form an Arithmetic Progression (AP).

The common difference $d$ is 3.

The last two-digit number divisible by 3 is 99.

---

So, the list of two-digit numbers divisible by 3 is:

$12, 15, 18, \dots, 99$.

This is an AP with:

First term, $a = 12$.

Common difference, $d = 3$.

Last term (or the $n$-th term), $a_n = 99$.

---

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

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

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

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

Subtract 12 from both sides:

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

$87 = 3(n-1)$

Divide both sides by 3:

Perform the division:

$29 = n-1$

Add 1 to both sides:

$n = 29 + 1$

$n = 30$

---

Since $n$ represents the number of terms in the AP, there are 30 two-digit numbers that are divisible by 3.

---

Conclusion: There are 30 two-digit numbers divisible by 3.

The given Arithmetic Progression (AP) is: $10, 7, 4, \dots, -62$.

---

The first term of this AP is $a = 10$.

The common difference $d$ is the difference between consecutive terms:

$d = 7 - 10 = -3$.

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

---

We need to find the 11th term from the last term of this AP.

One way to find the $k$-th term from the last of an AP is to use the formula $l - (k-1)d$, where $l$ is the last term and $d$ is the common difference. Here, $k=11$.

Using the formula for the $k$-th term from the end:

---

Substitute $l = -62$, $d = -3$, and $k = 11$ into formula (1):

$11$-th term from the end $= -62 - (11-1)(-3)$

$= -62 - (10)(-3)$

$= -62 - (-30)$

$= -62 + 30$

$= -32$

---

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

---

Alternate Solution:

Another way to solve this is to consider the AP in reverse order.

If we reverse the AP, the new first term ($a'$) is the last term of the original AP, and the new common difference ($d'$) is the negative of the original common difference.

Reversed AP: $-62, \dots, 4, 7, 10$.

First term of the reversed AP, $a' = -62$.

Common difference of the reversed AP, $d' = -d = -(-3) = 3$.

---

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

We use the formula for the $k$-th term of an AP, $a'_k = a' + (k-1)d'$, where $k$ is the term number we want (11th).

---

Substitute $a' = -62$, $d' = 3$, and $k = 11$ into formula (2):

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

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

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

$a'_{11} = -32$

---

Using both methods, we find that the 11th term from the last term of the AP is -32.

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

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

We need to calculate the simple interest at the end of each year.

The formula for simple interest for $T$ years is $I = \frac{P \times R \times T}{100}$.

---

Interest at the end of 1 year ($T=1$):

$I_1 = \frac{1000 \times 8 \times 1}{100} = \frac{8000}{100} = 80$.

Interest at the end of 2 years ($T=2$):

$I_2 = \frac{1000 \times 8 \times 2}{100} = \frac{16000}{100} = 160$.

Interest at the end of 3 years ($T=3$):

$I_3 = \frac{1000 \times 8 \times 3}{100} = \frac{24000}{100} = 240$.

And so on.

---

The interests at the end of 1, 2, 3, $\dots$ years are $80, 160, 240, \dots$.

Let's check if this list of numbers forms an AP.

Difference between 2nd and 1st term: $160 - 80 = 80$.

Difference between 3rd and 2nd term: $240 - 160 = 80$.

Since the difference between consecutive terms is constant (which is 80), the interests form an Arithmetic Progression (AP).

---

This AP is $80, 160, 240, \dots$.

The first term of this AP is $a = 80$.

The common difference is $d = 80$.

---

We need to find the interest at the end of 30 years. In this AP, the interest at the end of $n$ years is the $n$-th term ($a_n$). So, we need to find the 30th term ($a_{30}$) of this AP.

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

---

Substitute $a = 80$, $d = 80$, and $n = 30$ into formula (1):

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

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

$a_{30} = 80 + 2320$

$a_{30} = 2400$

---

The 30th term of the AP is 2400.

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

---

Conclusion: The annual interests form an AP. The interest at the end of 30 years is $\textsf{₹}2400$.

The number of rose plants in the rows are given as: $23, 21, 19, \dots, 5$.

---

Let's check if this sequence is an Arithmetic Progression (AP).

Difference between the second and first term: $21 - 23 = -2$.

Difference between the third and second term: $19 - 21 = -2$.

Since the difference between consecutive terms is constant, the sequence is an AP.

---

In this AP:

The first term is $a = 23$.

The common difference is $d = -2$.

The last term (let's call it the $n$-th term) is $a_n = 5$.

---

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

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

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

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

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 $n$ represents the number of terms (rows), there are 10 rows in the flower bed.

---

Conclusion: There are 10 rows in the flower bed.

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.

---

(i)

Given: $a = 7$, $d = 3$, $n = 8$.

To find: $a_n$ (specifically $a_8$).

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

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

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

$a_8 = 7 + 21$

$a_8 = 28$

The value of $a_n$ is $\mathbf{28}$.

---

(ii)

Given: $a = -18$, $n = 10$, $a_n = 0$.

To 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 both sides by 9:

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

$d = 2$

The value of $d$ is $\mathbf{2}$.

---

(iii)

Given: $d = -3$, $n = 18$, $a_n = -5$.

To find: $a$.

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

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

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

$-5 = a - 51$

Add 51 to both sides:

$a = -5 + 51$

$a = 46$

The value of $a$ is $\mathbf{46}$.

---

(iv)

Given: $a = -18.9$, $d = 2.5$, $a_n = 3.6$.

To find: $n$.

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

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

Add 18.9 to both sides:

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

$22.5 = (n-1)2.5$

Divide both sides by 2.5:

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

To simplify the division, multiply numerator and denominator by 10:

$n-1 = \frac{22.5 \times 10}{2.5 \times 10} = \frac{225}{25}$

$n-1 = 9$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

The value of $n$ is $\mathbf{10}$.

---

(v)

Given: $a = 3.5$, $d = 0$, $n = 105$.

To find: $a_n$ (specifically $a_{105}$).

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

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

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

$a_{105} = 3.5 + 0$

$a_{105} = 3.5$

The value of $a_n$ is $\mathbf{3.5}$.

To find the $n$-th term of an Arithmetic Progression (AP), we use the formula $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

---

(i) Given AP: $10, 7, 4, \dots$

The first term is $a = 10$.

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

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 $-77$.

Comparing with the given options, the correct choice is (C) –77.

---

(ii) Given AP: $-3, -\frac{1}{2}, 2, \dots$

The first term is $a = -3$.

The common difference $d$ is the difference between consecutive terms:

$d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = -\frac{1}{2} + \frac{6}{2} = \frac{5}{2}$.

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

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

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

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

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

$a_{11} = -3 + \frac{50}{2}$

$a_{11} = -3 + 25$

$a_{11} = 22$

The 11th term of the AP is $22$.

Comparing with the given options, the correct choice is (B) 22.

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

---

(i) Given AP: $2, \square, 26$

This AP has 3 terms. The first term is $a_1 = 2$ and the third term is $a_3 = 26$. The missing term is 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 $d = 12$.

The missing term is $a_2$, which is $a_1 + d$:

$a_2 = 2 + 12 = 14$.

The complete AP is $2, 14, 26$.

The missing term is $\mathbf{14}$.

---

(ii) Given AP: $\square, 13, \square, 3$

This AP has 4 terms. The second term is $a_2 = 13$ and the fourth term is $a_4 = 3$. The missing terms are the first term ($a_1$ or $a$) and the third term ($a_3$).

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

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

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

Subtract equation (1) from equation (2):

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

$2d = -10$

Divide by 2:

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

The common difference is $d = -5$.

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

$13 = a + (-5)$

$13 = a - 5$

$a = 13 + 5 = 18$.

The first term is $a_1 = 18$.

The missing third term is $a_3 = a_2 + d$ (or $a + 2d$):

$a_3 = 13 + (-5) = 13 - 5 = 8$.

The complete AP is $18, 13, 8, 3$.

The missing terms are $\mathbf{18}$ and $\mathbf{8}$.

---

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

This AP has 4 terms. The first term is $a_1 = 5$ and the fourth term is $a_4 = 9\frac{1}{2} = \frac{19}{2}$. The missing terms are the second ($a_2$) and third ($a_3$) terms.

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

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

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

The missing third term is $a_3 = a_2 + d$:

$a_3 = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8$.

The complete AP is $5, \frac{13}{2}, 8, \frac{19}{2}$.

The missing terms are $\mathbf{\frac{13}{2}}$ (or 6.5) and $\mathbf{8}$.

---

(iv) Given AP: – 4 , $\square$, $\square$, $\square$, $\square$, 6

This AP has 6 terms. The first term is $a_1 = -4$ and the sixth term is $a_6 = 6$. The missing terms are the second ($a_2$), third ($a_3$), fourth ($a_4$), and fifth ($a_5$) terms.

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 $d = 2$.

Now find the missing terms:

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

$a_3 = a_2 + d = -2 + 2 = 0$.

$a_4 = a_3 + d = 0 + 2 = 2$.

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

The complete AP is $-4, -2, 0, 2, 4, 6$.

The missing terms are $\mathbf{-2, 0, 2, 4}$.

---

(v) Given AP: $\square, 38, \square, \square, \square, – 22$

This AP has 6 terms. The second term is $a_2 = 38$ and the sixth term is $a_6 = -22$. The missing terms are the first ($a_1$ or $a$), third ($a_3$), fourth ($a_4$), and fifth ($a_5$) terms.

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

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

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

Subtract equation (1) from equation (2):

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

$4d = -60$

Divide by 4:

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

The common difference is $d = -15$.

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

$38 = a + (-15)$

$38 = a - 15$

$a = 38 + 15 = 53$.

The first term is $a_1 = 53$.

Now find the missing terms:

$a_3 = a_2 + d = 38 + (-15) = 38 - 15 = 23$.

$a_4 = a_3 + d = 23 + (-15) = 23 - 15 = 8$.

$a_5 = a_4 + d = 8 + (-15) = 8 - 15 = -7$.

The complete AP is $53, 38, 23, 8, -7, -22$.

The missing terms are $\mathbf{53, 23, 8, -7}$.

The given Arithmetic Progression (AP) is: $3, 8, 13, 18, \dots$

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 3 = 5$.

We can check this with the next pair: $13 - 8 = 5$. So, $d = 5$.

---

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

---

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

---

Substitute $a = 3$, $d = 5$, and $a_n = 78$ into formula (1):

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

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 a term in the AP. It is the 16th term.

---

Conclusion: The 16th term of the AP is 78.

To find the number of terms ($n$) in an Arithmetic Progression (AP), we use the formula for the $n$-th term: $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $a_n$ is the last term.

---

(i) Given AP: $7, 13, 19, \dots, 205$

Here, the first term $a = 7$.

The common difference $d = 13 - 7 = 6$. (Also, $19 - 13 = 6$)

The last term $a_n = 205$.

We need to find $n$.

Using 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:

$n = 33 + 1$

$n = 34$

The number of terms in the AP is 34.

---

(ii) Given AP: $18, 15\frac{1}{2}, 13, \dots, -47$

Here, the first term $a = 18$.

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

$15\frac{1}{2} = \frac{31}{2}$.

$d = \frac{31}{2} - 18 = \frac{31}{2} - \frac{36}{2} = \frac{31 - 36}{2} = -\frac{5}{2}$.

The last term $a_n = -47$.

We need to find $n$.

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

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

Subtract 18 from both sides:

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

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

Multiply both sides by $-\frac{2}{5}$:

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

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

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

$13 \times 2 = n-1$

$26 = n-1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

The number of terms in the AP is 27.

The given Arithmetic Progression (AP) is: $11, 8, 5, 2, \dots$

---

The first term of this AP is $a = 11$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 11 = -3$.

We can check this with the next pair: $5 - 8 = -3$. So, $d = -3$.

---

To check if $-150$ is a term of this AP, we need to determine if there exists a positive integer $n$ such that the $n$-th term, $a_n$, is equal to $-150$.

---

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

---

Substitute $a = 11$, $d = -3$, and $a_n = -150$ into formula (1):

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

Subtract 11 from both sides:

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

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

Divide both sides by -3:

Simplify the expression:

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

Add 1 to both sides to find $n$:

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

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

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

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

---

For $-150$ to be a term of the AP, the value of $n$ must be a positive integer.

The value we obtained for $n$ is $\frac{164}{3}$.

Since $\frac{164}{3}$ is not an integer (it is approximately $54.67$), $-150$ is not a term of the given AP.

---

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

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

---

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

---

We are given that the 11th term ($a_{11}$) is 38.

Using formula (1) with $n=11$:

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

$38 = a + 10d$

---

We are also given that the 16th term ($a_{16}$) is 73.

Using formula (1) with $n=16$:

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

$73 = a + 15d$

---

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

Subtract equation (i) from equation (ii) to eliminate $a$:

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

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

$5d = 35$

Divide by 5:

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

---

Substitute the value of $d=7$ into equation (i) to find $a$:

$a + 10(7) = 38$

$a + 70 = 38$

$a = 38 - 70$

---

Now that we have the first term ($a = -32$) and the common difference ($d = 7$), we can find the 31st term ($a_{31}$).

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 $\mathbf{178}$.

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

---

The AP consists of 50 terms, so the total number of terms is $N = 50$.

The last term is the 50th term, so $a_{50} = 106$.

We are given that the 3rd term is 12, so $a_3 = 12$.

---

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

---

Using formula (1) for the 3rd term ($n=3$):

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

$12 = a + 2d$

---

Using formula (1) for the last term (50th term, $n=50$):

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

$106 = a + 49d$

---

Now we have a system of two linear equations:

$a + 2d = 12$ (i)

$a + 49d = 106$ (ii)

Subtract equation (i) from equation (ii) to eliminate $a$:

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

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

$47d = 94$

Divide by 47:

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

---

Substitute the value of $d=2$ into equation (i) to find $a$:

$a + 2(2) = 12$

$a + 4 = 12$

$a = 12 - 4$

---

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 $n=29$:

$a_{29} = 8 + (29-1) \times 2$

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

$a_{29} = 8 + 56$

$a_{29} = 64$

---

Thus, the 29th term of the AP is $\mathbf{64}$.

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

---

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

---

We are given that the 3rd term ($a_3$) is 4.

Using formula (1) with $n=3$:

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

$4 = a + 2d$

---

We are also given that the 9th term ($a_9$) is -8.

Using formula (1) with $n=9$:

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

$-8 = a + 8d$

---

Now we have a system of two linear equations:

$a + 2d = 4$ (i)

$a + 8d = -8$ (ii)

Subtract equation (i) from equation (ii) to eliminate $a$:

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

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

$6d = -12$

Divide by 6:

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

---

Substitute the value of $d=-2$ into equation (i) to find $a$:

$a + 2(-2) = 4$

$a - 4 = 4$

$a = 4 + 4$

---

We need to find which term of this AP is zero. Let the $n$-th term be 0, i.e., $a_n = 0$.

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

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

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, 0 is a term in the AP. It is the 5th term.

---

Conclusion: The 5th term of the AP is zero.

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

---

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

---

The 17th term of the AP ($a_{17}$) is given by using formula (1) with $n=17$:

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

$a_{17} = a + 16d$

---

The 10th term of the AP ($a_{10}$) is given by using formula (1) with $n=10$:

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

$a_{10} = a + 9d$

---

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

---

Substitute the expressions for $a_{17}$ and $a_{10}$ from equations (i) and (ii) into the given relationship:

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

Simplify the equation:

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

Subtract $a$ from both sides:

$16d = 9d + 7$

Subtract $9d$ from both sides:

$16d - 9d = 7$

$7d = 7$

Divide both sides by 7:

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

$d = 1$

---

Thus, the common difference of the AP is $\mathbf{1}$.

The given Arithmetic Progression (AP) is: $3, 15, 27, 39, \dots$

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 15 - 3 = 12$.

We can verify this with the next pair: $27 - 15 = 12$. So, $d = 12$.

---

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

---

First, we find the 54th term of the AP. Using formula (1) with $n=54$:

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

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

$a_{54} = 3 + 636$

$a_{54} = 639$

---

Let the term which is 132 more than its 54th term be the $n$-th term, $a_n$.

According to the question:

$a_n = 639 + 132$

$a_n = 771$

---

Now, we need to find the value of $n$ for which $a_n = 771$.

Using the formula $a_n = a + (n-1)d$ again, with $a=3$, $d=12$, and $a_n=771$:

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

Subtract 3 from both sides:

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

$768 = 12(n-1)$

Divide both sides by 12:

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

Performing the division:

$64 = n-1$

Add 1 to both sides:

$n = 64 + 1$

$n = 65$

---

Since $n=65$ is a positive integer, 771 is a term in the AP. It is the 65th term.

---

Conclusion: The 65th term of the AP will be 132 more than its 54th term.

Let the first AP be denoted by $AP_1$ and the second AP by $AP_2$.

---

Let the first term of $AP_1$ be $a_1$ and its common difference be $d$.

Let the first term of $AP_2$ be $a_2$ and its common difference be $d'$.

According to the question, the two APs have the same common difference, so $d = d'$. Let's just use $d$ for the common difference for both APs.

---

The $n$-th term of $AP_1$ is given by $a_{1,n} = a_1 + (n-1)d$.

The $n$-th term of $AP_2$ is given by $a_{2,n} = a_2 + (n-1)d$.

---

We are given that the difference between their 100th terms is 100.

The 100th term of $AP_1$ is $a_{1,100} = a_1 + (100-1)d = a_1 + 99d$.

The 100th term of $AP_2$ is $a_{2,100} = a_2 + (100-1)d = a_2 + 99d$.

The difference between their 100th terms is:

Substitute the expressions for the terms:

$(a_1 + 99d) - (a_2 + 99d) = 100$

$a_1 + 99d - a_2 - 99d = 100$

$a_1 - a_2 = 100$

---

We need to find the difference between their 1000th terms.

The 1000th term of $AP_1$ is $a_{1,1000} = a_1 + (1000-1)d = a_1 + 999d$.

The 1000th term of $AP_2$ is $a_{2,1000} = a_2 + (1000-1)d = a_2 + 999d$.

The difference between their 1000th terms is:

$a_{1,1000} - a_{2,1000} = (a_1 + 999d) - (a_2 + 999d)$

$a_{1,1000} - a_{2,1000} = a_1 + 999d - a_2 - 999d$

$a_{1,1000} - a_{2,1000} = a_1 - a_2$

---

From equation (i), we know that $a_1 - a_2 = 100$.

Therefore, the difference between their 1000th terms is also 100.

$a_{1,1000} - a_{2,1000} = 100$

---

Conclusion: The difference between their 1000th terms is $\mathbf{100}$.

This is because the common difference cancels out when calculating the difference between the $n$-th terms of two APs with the same common difference, i.e., $(a_1 + (n-1)d) - (a_2 + (n-1)d) = a_1 - a_2$. The difference between any corresponding terms is equal to the difference between their first terms.

We are looking for the number of three-digit numbers that are divisible by 7.

The three-digit numbers range from 100 to 999.

---

To find the first three-digit number divisible by 7, we find the smallest number greater than or equal to 100 that is a multiple of 7.

Divide 100 by 7: $100 \div 7 \approx 14.28$.

The smallest integer multiple of 7 greater than 100 is $7 \times 15 = 105$.

So, the first three-digit number divisible by 7 is 105.

---

To find the last three-digit number divisible by 7, we find the largest number less than or equal to 999 that is a multiple of 7.

Divide 999 by 7: $999 \div 7 \approx 142.71$.

The largest integer multiple of 7 less than 999 is $7 \times 142 = 994$.

So, the last three-digit number divisible by 7 is 994.

---

The list of three-digit numbers divisible by 7 is:

$105, 112, 119, \dots, 994$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive terms is the common difference, which is 7 (since the numbers are multiples of 7).

---

In this AP:

The first term is $a = 105$.

The common difference is $d = 7$.

The last term (or the $n$-th term) is $a_n = 994$.

---

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

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

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

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

Subtract 105 from both sides:

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

$889 = 7(n-1)$

Divide both sides by 7:

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

Performing the division:

$127 = n-1$

Add 1 to both sides:

$n = 127 + 1$

$n = 128$

---

Since $n$ represents the number of terms in the AP, there are 128 three-digit numbers that are divisible by 7.

---

Conclusion: There are 128 three-digit numbers divisible by 7.

We are looking for the number of multiples of 4 that are strictly between 10 and 250. This means numbers greater than 10 and less than 250 that are divisible by 4.

---

To find the first multiple of 4 greater than 10, we check numbers starting from 11.

$11 \div 4 = 2$ with a remainder. $12 \div 4 = 3$.

The first multiple of 4 greater than 10 is 12.

---

To find the last multiple of 4 less than 250, we check numbers downwards from 249.

Divide 250 by 4: $250 \div 4 = 62$ with a remainder of 2.

So, $250 = 4 \times 62 + 2$.

To get a multiple of 4 less than 250, we subtract the remainder: $250 - 2 = 248$.

$248 \div 4 = 62$.

The last multiple of 4 less than 250 is 248.

---

The list of multiples of 4 between 10 and 250 is:

$12, 16, 20, \dots, 248$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive terms is the common difference, which is 4 (since the numbers are multiples of 4).

---

In this AP:

The first term is $a = 12$.

The common difference is $d = 4$.

The last term (or the $n$-th term) is $a_n = 248$.

---

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

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

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

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

Subtract 12 from both sides:

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

$236 = 4(n-1)$

Divide both sides by 4:

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

Performing the division:

$59 = n-1$

Add 1 to both sides:

$n = 59 + 1$

$n = 60$

---

Since $n$ represents the number of terms in the AP, there are 60 multiples of 4 between 10 and 250.

---

Conclusion: There are 60 multiples of 4 between 10 and 250.

Let the first Arithmetic Progression (AP) be $AP_1$ and the second AP be $AP_2$.

---

For $AP_1$: $63, 65, 67, \dots$

The first term is $a_1 = 63$.

The common difference is $d_1 = 65 - 63 = 2$.

The $n$-th term of $AP_1$ is given by the formula $a_{1,n} = a_1 + (n-1)d_1$.

---

For $AP_2$: $3, 10, 17, \dots$

The first term is $a_2 = 3$.

The common difference is $d_2 = 10 - 3 = 7$.

The $n$-th term of $AP_2$ is given by the formula $a_{2,n} = a_2 + (n-1)d_2$.

---

We need to find the value of $n$ for which the $n$-th terms of the two APs are equal, i.e., $a_{1,n} = a_{2,n}$.

Setting the expressions from equations (1) and (2) equal:

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

Expand both sides of the equation:

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

$61 + 2n = 7n - 4$

Rearrange the terms to solve for $n$. Subtract $2n$ from both sides and add 4 to both sides:

$61 + 4 = 7n - 2n$

$65 = 5n$

Divide both sides by 5:

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

$n = 13$

---

Since $n=13$ is a positive integer, the 13th terms of the two APs are equal.

Let's verify the 13th term for both APs:

For $AP_1$: $a_{1,13} = 63 + (13-1)2 = 63 + 12 \times 2 = 63 + 24 = 87$.

For $AP_2$: $a_{2,13} = 3 + (13-1)7 = 3 + 12 \times 7 = 3 + 84 = 87$.

The 13th terms are indeed equal (87).

---

Conclusion: The $n$-th terms of the two APs are equal when $n = \mathbf{13}$.

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

---

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

---

We are given that the 3rd term ($a_3$) is 16.

Using formula (1) with $n=3$:

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

$16 = a + 2d$

---

We are also given that the 7th term exceeds the 5th term by 12.

---

Using formula (1) for the 7th term ($n=7$):

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

---

Using formula (1) for the 5th term ($n=5$):

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

---

Substitute the expressions for $a_7$ and $a_5$ into the given condition $a_7 = a_5 + 12$:

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

Simplify the equation:

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

Subtract $a$ from both sides:

$6d = 4d + 12$

Subtract $4d$ from both sides:

$6d - 4d = 12$

$2d = 12$

Divide both sides by 2:

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

---

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

$a + 2(6) = 16$

$a + 12 = 16$

Subtract 12 from both sides:

$a = 16 - 12$

---

The Arithmetic Progression is determined by its first term ($a$) and its common difference ($d$).

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

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

Substituting the values of $a$ and $d$, the AP is:

$4, 4+6, 4+2(6), 4+3(6), \dots$

which is:

$4, 10, 16, 22, \dots$

---

The AP is $\mathbf{4, 10, 16, 22, \dots}$.

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

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 3 = 5$.

We can verify this with the next pair: $13 - 8 = 5$. So, $d = 5$.

---

The last term of the AP is $l = 253$.

---

We need to find the 20th term from the last term of this AP. Let $k=20$.

The formula for the $k$-th term from the last of an AP is given by:

---

Substitute the values $l = 253$, $d = 5$, and $k = 20$ into formula (1):

$20$-th term from the end $= 253 - (20-1) \times 5$

$= 253 - (19) \times 5$

$= 253 - 95$

$= 158$

---

Thus, the 20th term from the last term of the AP is $\mathbf{158}$.

---

Alternate Solution:

We can also find the total number of terms in the AP and then determine which term from the beginning corresponds to the 20th term from the last.

Let the number of terms in the AP be $n$. The last term is $a_n = 253$.

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

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

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

$250 = 5(n-1)$

$\frac{250}{5} = n-1$

$50 = n-1$

$n = 50 + 1 = 51$.

There are 51 terms in the AP.

---

The $k$-th term from the last of an AP with $n$ terms is the $(n - k + 1)$-th term from the beginning.

Here, $n=51$ and $k=20$.

The term number from the beginning is $51 - 20 + 1 = 31 + 1 = 32$.

So, the 20th term from the last is the 32nd term from the beginning ($a_{32}$).

---

Now, we find the 32nd term using the formula $a_n = a + (n-1)d$ with $a=3$, $d=5$, and $n=32$:

$a_{32} = 3 + (32-1) \times 5$

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

$a_{32} = 3 + 155$

$a_{32} = 158$

---

Using both methods, the 20th term from the last term of the AP is 158.

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

---

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

---

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

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

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

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

$2a + 10d = 24$

Divide the equation by 2:

---

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

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

The sum of the 6th and 10th terms is 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 $d$:

$a + 5d = 12$ (i)

$a + 7d = 22$ (ii)

Subtract equation (i) from equation (ii) to eliminate $a$:

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

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

$2d = 10$

Divide by 2:

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

---

Substitute the value of $d=5$ into equation (i) to find $a$:

$a + 5(5) = 12$

$a + 25 = 12$

Subtract 25 from both sides:

$a = 12 - 25$

---

Now that we have the first term ($a = -13$) and the common difference ($d = 5$), we can find the first three terms of the AP.

First term: $a_1 = a = -13$.

Second term: $a_2 = a + d = -13 + 5 = -8$.

Third term: $a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3$.

---

The first three terms of the AP are $\mathbf{-13, -8, -3}$.

Let the annual salary in the first year (1995) be the first term of an Arithmetic Progression (AP).

---

The initial salary in 1995 is $a_1 = \textsf{₹}5000$.

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

The annual salaries form an AP:

Year 1 (1995): $\textsf{₹}5000$ ($a_1$)

Year 2 (1996): $\textsf{₹}5000 + \textsf{₹}200 = \textsf{₹}5200$ ($a_2$)

Year 3 (1997): $\textsf{₹}5200 + \textsf{₹}200 = \textsf{₹}5400$ ($a_3$)

and so on.

The AP of salaries is: $5000, 5200, 5400, \dots$

---

We want to find the year in which his income reached $\textsf{₹}7000$. Let this salary be the $n$-th term of the AP, so $a_n = \textsf{₹}7000$.

---

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

---

Substitute the values $a = 5000$, $d = 200$, and $a_n = 7000$ into formula (1):

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

Subtract 5000 from both sides:

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

$2000 = 200(n-1)$

Divide both sides by 200:

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

$\frac{\cancel{2000}^{10}}{\cancel{200}_{1}} = n-1$

$10 = n-1$

Add 1 to both sides:

$n = 10 + 1$

$n = 11$

---

The $n$-th term being $\textsf{₹}7000$ means it is the 11th term of the AP.

The first term corresponds to the year 1995 ($n=1$).

The 11th term corresponds to the year $1995 + (11-1)$ years.

Year $= 1995 + 10 = 2005$.

---

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

Let the weekly savings in the first week be the first term of an Arithmetic Progression (AP).

---

Savings in the first week is $a_1 = \textsf{₹}5$.

The weekly increase in savings is constant, which represents the common difference, $d = \textsf{₹}1.75$.

The weekly savings form an AP:

Week 1: $\textsf{₹}5$ ($a_1$)

Week 2: $\textsf{₹}5 + \textsf{₹}1.75 = \textsf{₹}6.75$ ($a_2$)

Week 3: $\textsf{₹}6.75 + \textsf{₹}1.75 = \textsf{₹}8.50$ ($a_3$)

and so on.

The AP of weekly savings is: $5, 6.75, 8.50, \dots$

---

We are given that in the $n$-th week, her weekly savings become $\textsf{₹}20.75$. So, the $n$-th term is $a_n = \textsf{₹}20.75$.

---

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

---

Substitute the values $a = 5$, $d = 1.75$, and $a_n = 20.75$ into formula (1):

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

Subtract 5 from both sides:

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

$15.75 = 1.75(n-1)$

Divide both sides by 1.75:

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

To simplify the division, multiply the numerator and denominator by 100 to remove decimals:

$n-1 = \frac{15.75 \times 100}{1.75 \times 100} = \frac{1575}{175}$

We can perform the division:

$\frac{1575}{175} = \frac{315}{35} = \frac{63}{7} = 9$.

$n-1 = 9$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

---

Since $n=10$ is a positive integer, in the 10th week, Ramkali's weekly savings become $\textsf{₹}20.75$.

---

Conclusion: The value of $n$ is $\mathbf{10}$.

The given Arithmetic Progression (AP) is: $8, 3, -2, \dots$

---

Here, the first term is $a = 8$.

The common difference $d$ is the difference between consecutive terms:

$d = 3 - 8 = -5$.

We need to find the sum of the first 22 terms, so $n = 22$.

---

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

---

Substitute the values $a = 8$, $d = -5$, and $n = 22$ into formula (1) to find the sum of the first 22 terms ($S_{22}$):

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

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

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

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

So, $11 \times -89 = -979$.

$S_{22} = -979$

---

Thus, the sum of the first 22 terms of the AP is $\mathbf{-979}$.

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

---

We are given that the sum of the first 14 terms ($S_{14}$) is 1050, so $n=14$ and $S_{14} = 1050$.

The first term is $a = 10$.

---

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

---

Substitute $S_n = 1050$, $n = 14$, and $a = 10$ into formula (1):

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

$1050 = 7[20 + 13d]$

Divide both sides 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:

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

---

Now that we have the first term ($a = 10$) and the common difference ($d = 10$), we can find the 20th term ($a_{20}$).

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

Using the formula with $n=20$:

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

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

$a_{20} = 10 + 190$

$a_{20} = 200$

---

Thus, the 20th term of the AP is $\mathbf{200}$.

Given:

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

The sum of a certain number of terms of this AP is 78.

---

To Find:

The number of terms ($n$) that must be taken so that their sum is 78.

---

Solution:

For the given AP: 24, 21, 18, . . .

The first term, $a = 24$.

The common difference, $d = a_2 - a_1 = 21 - 24 = -3$.

The sum of $n$ terms, $S_n = 78$.

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

Substituting the given values into the formula:

Now, we solve this equation for $n$.

$78 = \frac{n}{2}[48 - 3n + 3]$

$78 = \frac{n}{2}[51 - 3n]$

Multiply both sides by 2:

$156 = n(51 - 3n)$

$156 = 51n - 3n^2$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$3n^2 - 51n + 156 = 0$

To simplify the equation, divide all terms by 3:

$n^2 - 17n + 52 = 0$

We can solve this quadratic equation by factoring. We need to find two numbers that multiply to 52 and add up to -17. These numbers are -4 and -13.

$n^2 - 4n - 13n + 52 = 0$

$n(n - 4) - 13(n - 4) = 0$

$(n - 4)(n - 13) = 0$

This gives two possible values for $n$:

$n - 4 = 0 \implies n = 4$

or

$n - 13 = 0 \implies n = 13$

Both values are positive integers, so both are possible.

Case 1: $n = 4$

The first 4 terms are 24, 21, 18, 15.

Sum = $24 + 21 + 18 + 15 = 78$.

Case 2: $n = 13$

The sum of the first 13 terms is also 78.

This happens because the AP is decreasing (common difference is negative). After a certain point, the terms become negative and cancel out some of the earlier positive terms.

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+9+6+3+0+(-3) \ $$ +(-6)+(-9)+(-12) = 0$.

Therefore, $S_{13} = S_4 + (\text{sum of terms from 5th to 13th}) = 78 + 0 = 78$.

Since the question asks how many terms must be taken, both answers are valid.

---

Hence, the sum of 4 terms or 13 terms of the given AP is 78.

The sequence of positive integers is $1, 2, 3, 4, \dots$. This sequence forms an Arithmetic Progression (AP) with the first term $a = 1$ and the common difference $d = 1$.

---

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

Alternatively, if the last term ($a_n$) is known, the sum can be calculated using:

---

(i) The first 1000 positive integers:

The sequence is $1, 2, 3, \dots, 1000$.

Here, the first term $a = 1$, the number of terms $n = 1000$, and the last term $a_n = 1000$.

Using formula (2) for the sum of the first 1000 terms ($S_{1000}$):

$S_{1000} = \frac{1000}{2}(1 + 1000)$

$S_{1000} = \cancel{\frac{1000}{2}}^{500}(1001)$

$S_{1000} = 500 \times 1001$

$S_{1000} = 500500$

The sum of the first 1000 positive integers is $\mathbf{500500}$.

---

(ii) The first n positive integers:

The sequence is $1, 2, 3, \dots, n$.

Here, the first term $a = 1$, the number of terms is $n$, and the last term $a_n = n$.

Using formula (2) for the sum of the first $n$ terms ($S_n$):

$S_n = \frac{n}{2}(1 + n)$

This is the well-known formula for the sum of the first $n$ positive integers.

The sum of the first $n$ positive integers is $\mathbf{\frac{n(n+1)}{2}}$.

The $n$-th term of the list of numbers is given by the formula:

---

We can find the first few terms of the sequence by substituting $n=1, 2, 3, \dots$ into the formula (i):

For $n=1$, the first term $a_1 = 3 + 2(1) = 3 + 2 = 5$.

For $n=2$, the second term $a_2 = 3 + 2(2) = 3 + 4 = 7$.

For $n=3$, the third term $a_3 = 3 + 2(3) = 3 + 6 = 9$.

The list of numbers is $5, 7, 9, \dots$.

---

Let's check if this sequence is an Arithmetic Progression (AP).

Difference between the second and first term: $a_2 - a_1 = 7 - 5 = 2$.

Difference between the third and second term: $a_3 - a_2 = 9 - 7 = 2$.

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

---

In this AP:

The first term is $a = a_1 = 5$.

The common difference is $d = 2$.

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 given by:

---

Substitute the values $a = 5$, $d = 2$, and $n = 24$ into formula (2) to find the sum of the first 24 terms ($S_{24}$):

$S_{24} = \frac{24}{2}[2(5) + (24-1)2]$

$S_{24} = 12[10 + (23)2]$

$S_{24} = 12[10 + 46]$

$S_{24} = 12[56]$

$S_{24} = 672$

---

Thus, the sum of the first 24 terms of the list of numbers is $\mathbf{672}$.

Let the production of TV sets in the first year be $a$.

Let the uniform increase in production each year (the common difference) be $d$.

The production in subsequent years forms an Arithmetic Progression (AP).

---

The production in the $n$-th year is given by the $n$-th term of the AP, $a_n = a + (n-1)d$.

---

We are given that the production in the third year is 600 sets. So, $a_3 = 600$.

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

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

$600 = a + 2d$

---

We are also given that the production in the seventh year is 700 sets. So, $a_7 = 700$.

Using the formula $a_n = a + (n-1)d$ with $n=7$:

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

$700 = a + 6d$

---

Now we have a system of two linear equations:

$a + 2d = 600$ (i)

$a + 6d = 700$ (ii)

Subtract equation (i) from equation (ii) to eliminate $a$:

$(a + 6d) - (a + 2d) = 700 - 600$

$a + 6d - a - 2d = 100$

$4d = 100$

Divide by 4:

$d = \frac{100}{4}$

---

Substitute the value of $d=25$ into equation (i) ($a + 2d = 600$) to find $a$:

$a + 2(25) = 600$

$a + 50 = 600$

Subtract 50 from both sides:

$a = 600 - 50$

---

(i) The production in the 1st year:

The production in the 1st year is the first term, $a$.

Production in 1st year = 550 sets.

---

(ii) The 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 $a=550$, $d=25$, and $n=10$:

$a_{10} = 550 + (10-1) \times 25$

$a_{10} = 550 + (9) \times 25$

$a_{10} = 550 + 225$

$a_{10} = 775$

The production in the 10th year was 775 sets.

---

(iii) The total production in first 7 years:

We need to find the sum of the first 7 terms of the AP, $S_7$.

We can use the formula for the sum of the first $n$ terms, $S_n = \frac{n}{2}(a + a_n)$, since we know the first term ($a=550$) and the 7th term ($a_7=700$). Here $n=7$.

$S_7 = \frac{7}{2}(a + a_7)$

$S_7 = \frac{7}{2}(550 + 700)$

$S_7 = \frac{7}{2}(1250)$

$S_7 = 7 \times \frac{1250}{2}$

$S_7 = 7 \times 625$

$S_7 = 4375$

The total production in the first 7 years was 4375 sets.

The formula for the sum of the first $n$ terms of an Arithmetic Progression (AP) is given by:

where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.

---

(i) Given AP: $2, 7, 12, \dots$, to 10 terms.

First term, $a = 2$.

Common difference, $d = 7 - 2 = 5$.

Number of terms, $n = 10$.

Using formula (1) to find the sum of the first 10 terms ($S_{10}$):

$S_{10} = \frac{10}{2}[2(2) + (10-1)5]$

$S_{10} = 5[4 + (9)5]$

$S_{10} = 5[4 + 45]$

$S_{10} = 5[49]$

$S_{10} = 245$

The sum of the first 10 terms is $\mathbf{245}$.

---

(ii) Given AP: $-37, -33, -29, \dots$, to 12 terms.

First term, $a = -37$.

Common difference, $d = -33 - (-37) = -33 + 37 = 4$.

Number of terms, $n = 12$.

Using formula (1) to find the sum of the first 12 terms ($S_{12}$):

$S_{12} = \frac{12}{2}[2(-37) + (12-1)4]$

$S_{12} = 6[-74 + (11)4]$

$S_{12} = 6[-74 + 44]$

$S_{12} = 6[-30]$

$S_{12} = -180$

The sum of the first 12 terms is $\mathbf{-180}$.

---

(iii) Given AP: $0.6, 1.7, 2.8, \dots$, to 100 terms.

First term, $a = 0.6$.

Common difference, $d = 1.7 - 0.6 = 1.1$.

Number of terms, $n = 100$.

Using formula (1) to find the sum of the first 100 terms ($S_{100}$):

$S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]$

$S_{100} = 50[1.2 + (99)1.1]$

Calculate $99 \times 1.1$: $99 \times (1 + 0.1) = 99 + 9.9 = 108.9$.

$S_{100} = 50[1.2 + 108.9]$

$S_{100} = 50[110.1]$

Calculate $50 \times 110.1$: $50 \times 110.1 = 5 \times 1101 = 5505$.

$S_{100} = 5505$

The sum of the first 100 terms is $\mathbf{5505}$.

---

(iv) Given AP: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms.

First term, $a = \frac{1}{15}$.

Common difference, $d = \frac{1}{12} - \frac{1}{15}$. The least common multiple of 12 and 15 is 60.

$d = \frac{1 \times 5}{12 \times 5} - \frac{1 \times 4}{15 \times 4} = \frac{5}{60} - \frac{4}{60} = \frac{1}{60}$.

Number of terms, $n = 11$.

Using formula (1) to find the sum of the first 11 terms ($S_{11}$):

$S_{11} = \frac{11}{2}[2(\frac{1}{15}) + (11-1)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + (10)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{10}{60}]$

Simplify $\frac{10}{60} = \frac{1}{6}$.

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}]$

The least common multiple of 15 and 6 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 $\frac{9}{30} = \frac{\cancel{9}^3}{\cancel{30}_{10}} = \frac{3}{10}$.

$S_{11} = \frac{11}{2} \times \frac{3}{10}$

$S_{11} = \frac{11 \times 3}{2 \times 10}$

$S_{11} = \frac{33}{20}$

The sum of the first 11 terms is $\mathbf{\frac{33}{20}}$.

To find the sum of an Arithmetic Progression (AP), we first need to identify the first term ($a$), the common difference ($d$), and the number of terms ($n$). If the last term ($a_n$) is known, we can use the formula $S_n = \frac{n}{2}(a + a_n)$. Otherwise, we use $S_n = \frac{n}{2}[2a + (n-1)d]$. To find $n$, we use the formula $a_n = a + (n-1)d$.

---

(i) Given series: $7 + 10\frac{1}{2} + 14 + \dots + 84$

This is an AP with:

First term, $a = 7$.

The common difference, $d = 10\frac{1}{2} - 7 = \frac{21}{2} - \frac{14}{2} = \frac{7}{2}$.

The last term is $a_n = 84$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$84 = 7 + (n-1)\frac{7}{2}$

$84 - 7 = (n-1)\frac{7}{2}$

$77 = (n-1)\frac{7}{2}$

Multiply both sides by $\frac{2}{7}$:

$77 \times \frac{2}{7} = n-1$

$\cancel{77}^{11} \times \frac{2}{\cancel{7}_1} = n-1$

$11 \times 2 = n-1$

$22 = n-1$

$n = 22 + 1 = 23$.

There are 23 terms in the AP.

Now find the sum of the first 23 terms ($S_{23}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{23} = \frac{23}{2}(7 + 84)$

$S_{23} = \frac{23}{2}(91)$

$S_{23} = \frac{23 \times 91}{2}$

$S_{23} = \frac{2093}{2}$

$S_{23} = 1046.5$ or $1046\frac{1}{2}$.

The sum of the given series is $\mathbf{1046.5}$ or $\mathbf{1046\frac{1}{2}}$.

---

(ii) Given series: $34 + 32 + 30 + \dots + 10$

This is an AP with:

First term, $a = 34$.

Common difference, $d = 32 - 34 = -2$.

The last term is $a_n = 10$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$10 = 34 + (n-1)(-2)$

$10 - 34 = (n-1)(-2)$

$-24 = -2(n-1)$

Divide both sides by -2:

$\frac{-24}{-2} = n-1$

$12 = n-1$

$n = 12 + 1 = 13$.

There are 13 terms in the AP.

Now find the sum of the first 13 terms ($S_{13}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{13} = \frac{13}{2}(34 + 10)$

$S_{13} = \frac{13}{2}(44)$

$S_{13} = 13 \times \frac{44}{2}$

$S_{13} = 13 \times 22$

$S_{13} = 286$

The sum of the given series is $\mathbf{286}$.

---

(iii) Given series: $-5 + (-8) + (-11) + \dots + (-230)$

This is an AP with:

First term, $a = -5$.

Common difference, $d = -8 - (-5) = -8 + 5 = -3$.

The last term is $a_n = -230$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$-230 = -5 + (n-1)(-3)$

$-230 - (-5) = (n-1)(-3)$

$-230 + 5 = (n-1)(-3)$

$-225 = -3(n-1)$

Divide both sides by -3:

$\frac{-225}{-3} = n-1$

$75 = n-1$

$n = 75 + 1 = 76$.

There are 76 terms in the AP.

Now find the sum of the first 76 terms ($S_{76}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{76} = \frac{76}{2}(-5 + (-230))$

$S_{76} = 38(-5 - 230)$

$S_{76} = 38(-235)$

Calculate $38 \times (-235)$: $38 \times 235 = 8930$. So $38 \times (-235) = -8930$.

The sum of the given series is $\mathbf{-8930}$.

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

The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}(a + a_n)$.

---

(i) Given: $a = 5$, $d = 3$, $a_n = 50$. Find $n$ and $S_n$.

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

$50 = 5 + (n-1)3$

$50 - 5 = 3(n-1)$

$45 = 3(n-1)$

$\frac{45}{3} = n-1$

$15 = n-1$

$n = 15 + 1 = 16$.

Now find $S_n$ (which is $S_{16}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{16} = \frac{16}{2}(5 + 50)$

$S_{16} = 8(55)$

$S_{16} = 440$.

So, $n = \mathbf{16}$ and $S_n = \mathbf{440}$.

---

(ii) Given: $a = 7$, $a_{13} = 35$. Find $d$ and $S_{13}$.

Using $a_n = a + (n-1)d$ with $n=13$ and $a_{13}=35$:

$35 = 7 + (13-1)d$

$35 = 7 + 12d$

$35 - 7 = 12d$

$28 = 12d$

$d = \frac{28}{12} = \frac{7}{3}$.

Now find $S_{13}$ using $S_n = \frac{n}{2}(a + a_n)$ with $n=13$ and $a_{13}=35$:

$S_{13} = \frac{13}{2}(7 + 35)$

$S_{13} = \frac{13}{2}(42)$

$S_{13} = 13 \times 21$

$S_{13} = 273$.

So, $d = \mathbf{\frac{7}{3}}$ and $S_{13} = \mathbf{273}$.

---

(iii) Given: $a_{12} = 37$, $d = 3$. Find $a$ and $S_{12}$.

Using $a_n = a + (n-1)d$ with $n=12$ and $a_{12}=37$:

$37 = a + (12-1)3$

$37 = a + 11 \times 3$

$37 = a + 33$

$a = 37 - 33 = 4$.

Now find $S_{12}$ using $S_n = \frac{n}{2}(a + a_n)$ with $n=12$ and $a_{12}=37$:

$S_{12} = \frac{12}{2}(4 + 37)$

$S_{12} = 6(41)$

$S_{12} = 246$.

So, $a = \mathbf{4}$ and $S_{12} = \mathbf{246}$.

---

(iv) Given: $a_3 = 15$, $S_{10} = 125$. Find $d$ and $a_{10}$.

Using $a_n = a + (n-1)d$ for $a_3$ ($n=3$):

Using $S_n = \frac{n}{2}[2a + (n-1)d]$ for $S_{10}$ ($n=10$):

$125 = \frac{10}{2}[2a + (10-1)d]$

$125 = 5[2a + 9d]$

Divide by 5:

$\frac{125}{5} = 2a + 9d$

From (i), $a = 15 - 2d$. Substitute this into (ii):

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

$25 - 30 = 5d$

$-5 = 5d$

$d = \frac{-5}{5} = -1$.

Now find $a$ using $a = 15 - 2d$:

$a = 15 - 2(-1) = 15 + 2 = 17$.

Finally, find $a_{10}$ using $a_n = a + (n-1)d$ with $a=17$, $d=-1$, and $n=10$:

$a_{10} = 17 + (10-1)(-1)$

$a_{10} = 17 + (9)(-1)$

$a_{10} = 17 - 9 = 8$.

So, $d = \mathbf{-1}$ and $a_{10} = \mathbf{8}$.

---

(v) Given: $d = 5$, $S_9 = 75$. Find $a$ and $a_9$.

Using $S_n = \frac{n}{2}[2a + (n-1)d]$ for $S_9$ ($n=9$):

$75 = \frac{9}{2}[2a + (9-1)5]$

$75 = \frac{9}{2}[2a + 8 \times 5]$

$75 = \frac{9}{2}[2a + 40]$

Multiply by 2 and divide by 9:

$\frac{75 \times 2}{9} = 2a + 40$

$\frac{150}{9} = 2a + 40$

$\frac{50}{3} = 2a + 40$

Subtract 40 from both sides:

$\frac{50}{3} - 40 = 2a$

$\frac{50}{3} - \frac{120}{3} = 2a$

$\frac{50 - 120}{3} = 2a$

$\frac{-70}{3} = 2a$

$a = \frac{-70}{3 \times 2} = \frac{-70}{6} = \frac{-35}{3}$.

Now find $a_9$ using $a_n = a + (n-1)d$ with $a = -\frac{35}{3}$, $d=5$, and $n=9$:

$a_9 = -\frac{35}{3} + (9-1)5$

$a_9 = -\frac{35}{3} + 8 \times 5$

$a_9 = -\frac{35}{3} + 40$

$a_9 = -\frac{35}{3} + \frac{120}{3}$

$a_9 = \frac{-35 + 120}{3} = \frac{85}{3}$.

So, $a = \mathbf{-\frac{35}{3}}$ and $a_9 = \mathbf{\frac{85}{3}}$.

---

(vi) Given: $a = 2$, $d = 8$, $S_n = 90$. Find $n$ and $a_n$.

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

$90 = \frac{n}{2}[2(2) + (n-1)8]$

$90 = \frac{n}{2}[4 + 8n - 8]$

$90 = \frac{n}{2}[8n - 4]$

$90 = n(4n - 2)$

$90 = 4n^2 - 2n$

Rearrange into a quadratic equation:

$4n^2 - 2n - 90 = 0$

Divide by 2:

$2n^2 - n - 45 = 0$.

Use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-1$, $c=-45$.

$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-45)}}{2(2)}$

$n = \frac{1 \pm \sqrt{1 + 360}}{4}$

$n = \frac{1 \pm \sqrt{361}}{4}$

$n = \frac{1 \pm 19}{4}$.

Two possible values for $n$:

$n_1 = \frac{1 + 19}{4} = \frac{20}{4} = 5$

$n_2 = \frac{1 - 19}{4} = \frac{-18}{4} = -\frac{9}{2}$.

Since $n$ must be a positive integer, $n=5$ is the only valid solution.

Now find $a_n$ (which is $a_5$) using $a_n = a + (n-1)d$ with $a=2$, $d=8$, and $n=5$:

$a_5 = 2 + (5-1)8$

$a_5 = 2 + 4 \times 8$

$a_5 = 2 + 32 = 34$.

So, $n = \mathbf{5}$ and $a_n = \mathbf{34}$.

---

(vii) Given: $a = 8$, $a_n = 62$, $S_n = 210$. Find $n$ and $d$.

Using $S_n = \frac{n}{2}(a + a_n)$:

$210 = \frac{n}{2}(8 + 62)$

$210 = \frac{n}{2}(70)$

$210 = 35n$

$n = \frac{210}{35} = 6$.

Now use $a_n = a + (n-1)d$ with $a=8$, $a_n=62$, and $n=6$ to find $d$:

$62 = 8 + (6-1)d$

$62 = 8 + 5d$

$62 - 8 = 5d$

$54 = 5d$

$d = \frac{54}{5}$.

So, $n = \mathbf{6}$ and $d = \mathbf{\frac{54}{5}}$ (or 10.8).

---

(viii) Given: $a_n = 4$, $d = 2$, $S_n = –14$. Find $n$ and $a$.

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

Using $S_n = \frac{n}{2}(a + a_n)$:

$-14 = \frac{n}{2}(a + 4)$

Multiply by 2:

From (i), $4 = a + 2n - 2 \implies 6 = a + 2n \implies a = 6 - 2n$.

Substitute this expression for $a$ into equation (ii):

$-28 = n((6 - 2n) + 4)$

$-28 = n(10 - 2n)$

$-28 = 10n - 2n^2$

Rearrange into a quadratic equation:

$2n^2 - 10n - 28 = 0$

Divide by 2:

$n^2 - 5n - 14 = 0$.

Factorize the quadratic equation. We look for two numbers that multiply to -14 and add up to -5. These are -7 and 2.

$(n - 7)(n + 2) = 0$.

Two possible values for $n$:

$n - 7 = 0 \implies n = 7$

$n + 2 = 0 \implies n = -2$.

Since $n$ must be a positive integer, $n=7$ is the only valid solution.

Now find $a$ using $a = 6 - 2n$ with $n=7$:

$a = 6 - 2(7)$

$a = 6 - 14 = -8$.

So, $n = \mathbf{7}$ and $a = \mathbf{-8}$.

---

(ix) Given: $a = 3$, $n = 8$, $S = 192$. Find $d$.

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

$192 = \frac{8}{2}[2(3) + (8-1)d]$

$192 = 4[6 + 7d]$

Divide by 4:

$\frac{192}{4} = 6 + 7d$

$48 = 6 + 7d$

$48 - 6 = 7d$

$42 = 7d$

$d = \frac{42}{7} = 6$.

So, $d = \mathbf{6}$.

---

(x) Given: $l = 28$, $S = 144$, and there are total 9 terms. Find $a$.

The last term is given as $l$, which is the $n$-th term, $a_n$. So, $a_n = 28$.

The total number of terms is 9, so $n = 9$.

The sum is given as $S = 144$, so $S_9 = 144$.

We need to find the first term, $a$.

Using the formula $S_n = \frac{n}{2}(a + a_n)$, substitute $S_9=144$, $n=9$, and $a_n=28$:

$144 = \frac{9}{2}(a + 28)$

Multiply by 2 and divide by 9:

$\frac{144 \times 2}{9} = a + 28$

$\frac{288}{9} = a + 28$

$32 = a + 28$

Subtract 28 from both sides:

$a = 32 - 28 = 4$.

So, $a = \mathbf{4}$.

The given Arithmetic Progression (AP) is: $9, 17, 25, \dots$

---

Here, the first term is $a = 9$.

The common difference $d$ is the difference between consecutive terms:

$d = 17 - 9 = 8$.

We want to find the number of terms ($n$) that must be taken so that their sum ($S_n$) is 636.

$S_n = 636$.

---

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

---

Substitute $S_n = 636$, $a = 9$, and $d = 8$ into formula (1):

$636 = \frac{n}{2}[2(9) + (n-1)8]$

$636 = \frac{n}{2}[18 + 8n - 8]$

$636 = \frac{n}{2}[10 + 8n]$

$636 = n(5 + 4n)$

$636 = 5n + 4n^2$

Rearrange the terms to form a quadratic equation:

$4n^2 + 5n - 636 = 0$

---

Now we solve this quadratic equation for $n$. We can use factorization or the quadratic formula.

Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=4$, $b=5$, $c=-636$.

First, calculate the discriminant $\Delta = b^2 - 4ac$:

$\Delta = (5)^2 - 4(4)(-636)$

$\Delta = 25 + 16(636)$

$\Delta = 25 + 10176$

$\Delta = 10201$

Now, find the square root of the discriminant:

$\sqrt{10201}$. We know $100^2 = 10000$, so the number is close to 100 and ends in 1. It could be 101.

$101 \times 101 = 10201$.

So, $\sqrt{10201} = 101$.

---

Now substitute the values into the quadratic formula for $n$:

$n = \frac{-5 \pm 101}{2(4)}$

$n = \frac{-5 \pm 101}{8}$.

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 $n$ represents the number of terms, it must be a positive integer. Therefore, $n=12$ is the only valid solution.

---

Conclusion: $\mathbf{12}$ terms of the AP must be taken to give a sum of 636.

Given: First term $a = 5$.

Given: Last term $a_n = 45$.

Given: Sum of the terms $S_n = 400$.

---

We need to find the number of terms ($n$) and the common difference ($d$).

---

We can use the formula for the sum of the first $n$ terms when the first and last terms are known:

---

Substitute the given values into formula (1):

$400 = \frac{n}{2}(5 + 45)$

$400 = \frac{n}{2}(50)$

$400 = 25n$

Divide both sides by 25:

$n = \frac{400}{25}$

$n = 16$.

---

So, the number of terms in the AP is $\mathbf{16}$.

---

Now that we know $n=16$, the last term $a_n=45$ is actually the 16th term, $a_{16}=45$.

We can use the formula for the $n$-th term of an AP to find the common difference $d$:

---

Substitute $a_{16} = 45$, $a = 5$, and $n = 16$ into formula (2):

$45 = 5 + (16-1)d$

$45 = 5 + 15d$

Subtract 5 from both sides:

$45 - 5 = 15d$

$40 = 15d$

Divide both sides by 15:

$d = \frac{40}{15}$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, 5:

$d = \frac{40 \div 5}{15 \div 5} = \frac{8}{3}$.

---

So, the common difference is $\mathbf{\frac{8}{3}}$.

Given: First term $a = 17$.

Given: Last term $a_n = 350$.

Given: Common difference $d = 9$.

---

We need to find the number of terms ($n$) and the sum of the terms ($S_n$).

---

First, find the number of terms ($n$) using the formula for the $n$-th term of an AP:

---

Substitute the given values into formula (1):

$350 = 17 + (n-1)9$

Subtract 17 from both sides:

$350 - 17 = (n-1)9$

$333 = 9(n-1)$

Divide both sides by 9:

$\frac{333}{9} = n-1$

Performing the division:

$37 = n-1$

Add 1 to both sides:

$n = 37 + 1 = 38$.

---

So, the number of terms is $\mathbf{38}$.

---

Now, find the sum of these 38 terms ($S_{38}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

---

Substitute the values $n=38$, $a=17$, and $a_{38}=350$ into formula (2):

$S_{38} = \frac{38}{2}(17 + 350)$

$S_{38} = 19(367)$

$S_{38} = 6973$.

---

Thus, the sum of the 38 terms is $\mathbf{6973}$.

Given: Common difference $d = 7$.

Given: The 22nd term $a_{22} = 149$.

We need to find the sum of the first 22 terms, $S_{22}$. This means the number of terms is $n=22$.

---

To find the sum of the first 22 terms, we can use the formula $S_n = \frac{n}{2}(a + a_n)$, since we know $n$, $d$, and $a_n$ (where $n=22$). However, we first need to find the first term ($a$).

---

Use the formula for the $n$-th term of an AP to find the first term ($a$):

---

Substitute $a_{22} = 149$, $n = 22$, and $d = 7$ into formula (1):

$149 = a + (22-1)7$

$149 = a + (21)7$

$149 = a + 147$

Subtract 147 from both sides:

$a = 149 - 147$

$a = 2$.

---

So, the first term is $a = 2$.

---

Now, find the sum of the first 22 terms ($S_{22}$) using the formula $S_n = \frac{n}{2}(a + a_n)$ with $n=22$, $a=2$, and $a_{22}=149$:

---

$S_{22} = \frac{22}{2}(2 + 149)$

$S_{22} = 11(151)$

$S_{22} = 1661$.

---

Thus, the sum of the first 22 terms of the AP is $\mathbf{1661}$.

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

---

We are given that the second term ($a_2$) is 14, and the third term ($a_3$) is 18.

---

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

$d = a_3 - a_2$

$d = 18 - 14$

$d = 4$.

---

Now we need to find the first term ($a$). The second term is $a_2 = a + d$.

$14 = a + 4$

Subtract 4 from both sides:

$a = 14 - 4$

$a = 10$.

---

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, so $n = 51$.

---

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

---

Substitute the values $a = 10$, $d = 4$, and $n = 51$ into formula (1) to find the sum of the first 51 terms ($S_{51}$):

$S_{51} = \frac{51}{2}[2(10) + (51-1)4]$

$S_{51} = \frac{51}{2}[20 + (50)4]$

$S_{51} = \frac{51}{2}[20 + 200]$

$S_{51} = \frac{51}{2}[220]$

$S_{51} = 51 \times \frac{220}{2}$

$S_{51} = 51 \times 110$

$51 \times 11 = 561$.

So, $S_{51} = 561 \times 10 = 5610$.

---

Thus, the sum of the first 51 terms of the AP is $\mathbf{5610}$.

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

---

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

---

We are given that the sum of the first 7 terms ($S_7$) is 49.

Using formula (1) with $n=7$ and $S_7=49$:

$49 = \frac{7}{2}[2a + (7-1)d]$

$49 = \frac{7}{2}[2a + 6d]$

$49 = \frac{7}{2} \times 2[a + 3d]$

$49 = 7[a + 3d]$

Divide both sides by 7:

$\frac{49}{7} = a + 3d$

---

We are also given that the sum of the first 17 terms ($S_{17}$) is 289.

Using formula (1) with $n=17$ and $S_{17}=289$:

$289 = \frac{17}{2}[2a + (17-1)d]$

$289 = \frac{17}{2}[2a + 16d]$

$289 = \frac{17}{2} \times 2[a + 8d]$

$289 = 17[a + 8d]$

Divide both sides by 17 ($289 = 17^2$, so $289 \div 17 = 17$):

$\frac{289}{17} = a + 8d$

---

Now we have a system of two linear equations:

$a + 3d = 7$ (i)

$a + 8d = 17$ (ii)

Subtract equation (i) from equation (ii) to eliminate $a$:

$(a + 8d) - (a + 3d) = 17 - 7$

$a + 8d - a - 3d = 10$

$5d = 10$

Divide by 5:

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

---

Substitute the value of $d=2$ into equation (i) ($a + 3d = 7$) to find $a$:

$a + 3(2) = 7$

$a + 6 = 7$

$a = 7 - 6$

---

Now that we have the first term ($a = 1$) and the common difference ($d = 2$), we can find the sum of the first $n$ terms ($S_n$).

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

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

$S_n = \frac{n}{2}[2 + 2n - 2]$

$S_n = \frac{n}{2}[2n]$

$S_n = \frac{n}{\cancel{2}_1} \times \cancel{2}^1 n$

$S_n = n \times n$

$S_n = n^2$

---

Thus, the sum of the first $n$ terms of the AP is $\mathbf{n^2}$.

A sequence $a_1, a_2, a_3, \dots, a_n, \dots$ forms an Arithmetic Progression (AP) if the difference between consecutive terms is constant. That is, $a_{n+1} - a_n = d$ for all $n \geq 1$, where $d$ is the common difference.

---

(i) Given $a_n = 3 + 4n$.

To show it is an AP, we find the difference between the $(n+1)$-th term and the $n$-th term.

$a_{n+1} = 3 + 4(n+1) = 3 + 4n + 4 = 7 + 4n$.

$a_{n+1} - a_n = (7 + 4n) - (3 + 4n)$

$a_{n+1} - a_n = 7 + 4n - 3 - 4n$

$a_{n+1} - a_n = 4$.

Since the difference between consecutive terms is a constant (4), the sequence forms an AP with common difference $d = 4$.

---

Now find the sum of the first 15 terms ($S_{15}$).

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 is $n=15$.

Using the 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 $\mathbf{525}$.

---

(ii) Given $a_n = 9 – 5n$.

To show it is an AP, we find the difference between the $(n+1)$-th term and the $n$-th term.

$a_{n+1} = 9 - 5(n+1) = 9 - 5n - 5 = 4 - 5n$.

$a_{n+1} - a_n = (4 - 5n) - (9 - 5n)$

$a_{n+1} - a_n = 4 - 5n - 9 + 5n$

$a_{n+1} - a_n = -5$.

Since the difference between consecutive terms is a constant (-5), the sequence forms an AP with common difference $d = -5$.

---

Now find the sum of the first 15 terms ($S_{15}$).

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 is $n=15$.

Using the 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 $\mathbf{-465}$.

The sum of the first $n$ terms of an AP is given by $S_n = 4n - n^2$.

---

The first term ($a_1$) is the sum of the first 1 term ($S_1$).

$S_1 = 4(1) - (1)^2 = 4 - 1 = 3$.

So, the first term is $a_1 = S_1 = \mathbf{3}$.

---

The sum of the first two terms ($S_2$) is found by substituting $n=2$ into the formula for $S_n$.

$S_2 = 4(2) - (2)^2 = 8 - 4 = 4$.

The sum of the first two terms is $\mathbf{4}$.

---

The second term ($a_2$) is the difference between the sum of the first two terms ($S_2$) and the sum of the first one term ($S_1$).

$a_2 = S_2 - S_1$

$a_2 = 4 - 3 = 1$.

The second term is $\mathbf{1}$.

---

The common difference ($d$) of the AP is the difference between the second term and the first term.

$d = a_2 - a_1 = 1 - 3 = -2$.

---

The third term ($a_3$) can be found using $a_3 = a_2 + d$ or using the relationship $a_n = S_n - S_{n-1}$ for $n > 1$.

Using $a_3 = a_2 + d$: $a_3 = 1 + (-2) = 1 - 2 = -1$.

Alternatively, using $a_n = S_n - S_{n-1}$: We find $S_3$.

$S_3 = 4(3) - (3)^2 = 12 - 9 = 3$.

$a_3 = S_3 - S_2 = 3 - 4 = -1$.

The third term is $\mathbf{-1}$.

---

The 10th term ($a_{10}$) can be found using the formula $a_n = a_1 + (n-1)d$, with $a_1 = 3$ and $d = -2$, and $n=10$.

$a_{10} = 3 + (10-1)(-2)$

$a_{10} = 3 + (9)(-2)$

$a_{10} = 3 - 18 = -15$.

The 10th term is $\mathbf{-15}$.

---

The $n$-th term ($a_n$) can be found using the formula $a_n = a_1 + (n-1)d$ with $a_1 = 3$ and $d = -2$.

$a_n = 3 + (n-1)(-2)$

$a_n = 3 - 2n + 2$

$a_n = 5 - 2n$.

Alternatively, using $a_n = S_n - S_{n-1}$ for $n > 1$:

$S_{n-1} = 4(n-1) - (n-1)^2$

$S_{n-1} = 4n - 4 - (n^2 - 2n + 1)$

$S_{n-1} = 4n - 4 - n^2 + 2n - 1$

$S_{n-1} = -n^2 + 6n - 5$.

$a_n = S_n - S_{n-1}$

$a_n = (4n - n^2) - (-n^2 + 6n - 5)$

$a_n = 4n - n^2 + n^2 - 6n + 5$

$a_n = -2n + 5 = 5 - 2n$.

This formula holds for $n > 1$. For $n=1$, $a_1 = 5 - 2(1) = 3$, which matches $S_1$. So the formula $a_n = 5-2n$ is valid for all positive integers $n$.

The $n$-th term is $\mathbf{5-2n}$.

The positive integers divisible by 6 are the multiples of 6.

The first few positive integers divisible by 6 are $6, 12, 18, 24, \dots$.

This sequence forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 6$.

The common difference is $d = 12 - 6 = 6$.

We need to find the sum of the first 40 terms, so $n = 40$.

---

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

---

Substitute the values $a = 6$, $d = 6$, and $n = 40$ into formula (1) to find the sum of the first 40 terms ($S_{40}$):

$S_{40} = \frac{40}{2}[2(6) + (40-1)6]$

$S_{40} = 20[12 + (39)6]$

Calculate $39 \times 6$:

$39 \times 6 = (40 - 1) \times 6 = 40 \times 6 - 1 \times 6 = 240 - 6 = 234$.

$S_{40} = 20[12 + 234]$

$S_{40} = 20[246]$

Calculate $20 \times 246$:

$20 \times 246 = 2 \times 10 \times 246 = 2 \times 2460 = 4920$.

$S_{40} = 4920$.

---

Thus, the sum of the first 40 positive integers divisible by 6 is $\mathbf{4920}$.

---

Alternate Method:

The first 40 positive integers divisible by 6 are $6 \times 1, 6 \times 2, 6 \times 3, \dots, 6 \times 40$.

The sequence is $6, 12, 18, \dots, 240$.

This is an AP with $a = 6$, $n = 40$, and the last term $a_{40} = 240$.

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{40} = \frac{40}{2}(6 + 240)$

$S_{40} = 20(246)$

$S_{40} = 4920$.

The sum is 4920.

The first 15 multiples of 8 are $8 \times 1, 8 \times 2, 8 \times 3, \dots, 8 \times 15$.

The sequence is $8, 16, 24, \dots, 120$.

This sequence forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 8$.

The common difference is $d = 16 - 8 = 8$.

The number of terms is $n = 15$.

The last term is $a_{15} = 8 \times 15 = 120$.

---

We need to find the sum of the first 15 terms, $S_{15}$.

Using the formula for the sum of the first $n$ terms when the first and last terms are known:

---

Substitute the values $n=15$, $a=8$, and $a_{15}=120$ into formula (1):

$S_{15} = \frac{15}{2}(8 + 120)$

$S_{15} = \frac{15}{2}(128)$

$S_{15} = 15 \times \frac{128}{2}$

$S_{15} = 15 \times 64$

$S_{15} = 960$.

---

Thus, the sum of the first 15 multiples of 8 is $\mathbf{960}$.

---

Alternate Method:

Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $a=8$, $d=8$, and $n=15$:

$S_{15} = \frac{15}{2}[2(8) + (15-1)8]$

$S_{15} = \frac{15}{2}[16 + (14)8]$

$S_{15} = \frac{15}{2}[16 + 112]$

$S_{15} = \frac{15}{2}[128]$

$S_{15} = 15 \times 64 = 960$.

The sum is 960.

The odd numbers between 0 and 50 are $1, 3, 5, \dots, 49$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive odd numbers is constant (2).

---

In this AP:

The first term is $a = 1$.

The common difference is $d = 3 - 1 = 2$.

The last term is $a_n = 49$.

---

First, we need to find the number of terms ($n$) in this AP.

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

---

Substitute the values $a=1$, $d=2$, and $a_n=49$ into formula (1):

$49 = 1 + (n-1)2$

$49 - 1 = (n-1)2$

$48 = 2(n-1)$

Divide both sides by 2:

$\frac{48}{2} = n-1$

$24 = n-1$

Add 1 to both sides:

$n = 24 + 1 = 25$.

---

So, there are 25 odd numbers between 0 and 50.

---

Now, find the sum of these 25 terms ($S_{25}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

---

Substitute the values $n=25$, $a=1$, and $a_{25}=49$ into formula (2):

$S_{25} = \frac{25}{2}(1 + 49)$

$S_{25} = \frac{25}{2}(50)$

$S_{25} = 25 \times \frac{50}{2}$

$S_{25} = 25 \times 25$

$S_{25} = 625$.

---

Thus, the sum of the odd numbers between 0 and 50 is $\mathbf{625}$.

The penalties for delay for each day form a sequence:

Day 1 penalty: $\textsf{₹}200$

Day 2 penalty: $\textsf{₹}250$

Day 3 penalty: $\textsf{₹}300$

and so on.

---

The penalty for each succeeding day is $\textsf{₹}50$ more than for the preceding day. This means the sequence of penalties forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 200$.

The common difference is $d = 250 - 200 = 50$. (Also, $300 - 250 = 50$).

The contractor delayed the work by 30 days, so we need to find the total penalty for 30 days. This means we need to find the sum of the first 30 terms of this AP, i.e., $n=30$ and we need to calculate $S_{30}$.

---

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

---

Substitute the values $a = 200$, $d = 50$, and $n = 30$ into formula (1) to find the total penalty for 30 days ($S_{30}$):

$S_{30} = \frac{30}{2}[2(200) + (30-1)50]$

$S_{30} = 15[400 + (29)50]$

Calculate $29 \times 50$: $29 \times 50 = 29 \times 5 \times 10 = 145 \times 10 = 1450$.

$S_{30} = 15[400 + 1450]$

$S_{30} = 15[1850]$

$S_{30} = 27750$.

---

The total money the contractor has to pay as penalty for delaying the work by 30 days is $\textsf{₹}27750$.

---

Conclusion: The contractor has to pay a penalty of $\mathbf{\textsf{₹}27750}$.

There are seven cash prizes, so the number of prizes is $n=7$.

The total sum of the prizes is $\textsf{₹}700$. This is the sum of the first 7 terms of the sequence of prize values, $S_7 = 700$.

---

Each prize is $\textsf{₹}20$ less than its preceding prize. This means the sequence of prize values forms an Arithmetic Progression (AP).

Let the value of the first prize be $a$ (the first term). The common difference is $d = -20$ (since the value decreases). The prizes are $a, a-20, a-40, \dots$.

---

We need to find the value of each of the seven prizes. This means finding the terms $a_1, a_2, \dots, a_7$. First, we need to find the first term $a$ and the common difference $d$. We already know $d=-20$.

---

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

---

Substitute $S_7 = 700$, $n = 7$, and $d = -20$ into formula (1):

$700 = \frac{7}{2}[2a + (7-1)(-20)]$

$700 = \frac{7}{2}[2a + (6)(-20)]$

$700 = \frac{7}{2}[2a - 120]$

$700 = \frac{7}{\cancel{2}} \times \cancel{2}[a - 60]$

$700 = 7(a - 60)$

Divide both sides by 7:

$\frac{700}{7} = a - 60$

$100 = a - 60$

Add 60 to both sides:

$a = 100 + 60$

$a = 160$.

---

The first term (the value of the first prize) is $a = \textsf{₹}160$.

The common difference is $d = -20$.

---

Now we can find the value of each of the seven prizes:

1st Prize ($a_1$): $a = 160$.

2nd Prize ($a_2$): $a + d = 160 + (-20) = 160 - 20 = 140$.

3rd Prize ($a_3$): $a + 2d = 160 + 2(-20) = 160 - 40 = 120$.

4th Prize ($a_4$): $a + 3d = 160 + 3(-20) = 160 - 60 = 100$.

5th Prize ($a_5$): $a + 4d = 160 + 4(-20) = 160 - 80 = 80$.

6th Prize ($a_6$): $a + 5d = 160 + 5(-20) = 160 - 100 = 60$.

7th Prize ($a_7$): $a + 6d = 160 + 6(-20) = 160 - 120 = 40$.

---

The values of the seven prizes are $\textsf{₹}160, \textsf{₹}140, \textsf{₹}120, \textsf{₹}100, \textsf{₹}80, \textsf{₹}60$, and $\textsf{₹}40$.

Let's check the sum: $160 + 140 + 120 + 100 + 80 + 60 + 40 \ $$ = 300 + 220 + 140 + 40 = 520 + 140 + 40 \ $$ = 660 + 40 = 700$. The sum is correct.

---

Conclusion: The values of the prizes are $\textsf{₹}160, \textsf{₹}140, \textsf{₹}120, \textsf{₹}100, \textsf{₹}80, \textsf{₹}60, \textsf{₹}40$.

The number of trees planted by each section is equal to the class number. There are 12 classes (Class I to Class XII).

There are 3 sections for each class.

---

Number of trees planted by each section of Class I = 1 tree.

Total trees planted by Class I = Number of sections $\times$ Trees per section = $3 \times 1 = 3$ trees.

---

Number of trees planted by each section of Class II = 2 trees.

Total trees planted by Class II = Number of sections $\times$ Trees per section = $3 \times 2 = 6$ trees.

---

Number of trees planted by each section of Class III = 3 trees.

Total trees planted by Class III = Number of sections $\times$ Trees per section = $3 \times 3 = 9$ trees.

---

... and so on, up to Class XII.

---

Number of trees planted by each section of Class XII = 12 trees.

Total trees planted by Class XII = Number of sections $\times$ Trees per section = $3 \times 12 = 36$ trees.

---

The total number of trees planted by the students is the sum of the trees planted by each class:

Total trees = (Trees by Class I) + (Trees by Class II) + ... + (Trees by Class XII)

Total trees = $3 + 6 + 9 + \dots + 36$.

---

This sequence of numbers $3, 6, 9, \dots, 36$ forms an Arithmetic Progression (AP).

In this AP:

The first term is $a = 3$.

The common difference is $d = 6 - 3 = 3$. (Also, $9 - 6 = 3$).

The last term is $a_{12} = 36$.

The number of terms is $n = 12$ (since there are 12 classes).

---

We need to find the sum of this AP ($S_{12}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

---

Substitute the values $n=12$, $a=3$, and $a_{12}=36$ into formula (1):

$S_{12} = \frac{12}{2}(3 + 36)$

$S_{12} = 6(39)$

Calculate $6 \times 39$:

$6 \times 39 = 6 \times (40 - 1) = 240 - 6 = 234$.

$S_{12} = 234$.

---

Thus, the total number of trees planted by the students is 234.

---

Conclusion: The total number of trees planted will be $\mathbf{234}$.

The spiral is made up of successive semicircles with radii $r_1 = 0.5$ cm, $r_2 = 1.0$ cm, $r_3 = 1.5$ cm, $r_4 = 2.0$ cm, and so on.

---

The length of a semicircle with radius $r$ is given by $\pi r$.

The lengths of the successive semicircles are:

Length of the first semicircle ($l_1$) with radius $r_1 = 0.5$ cm $= \pi r_1 = \pi(0.5)$ cm.

Length of the second semicircle ($l_2$) with radius $r_2 = 1.0$ cm $= \pi r_2 = \pi(1.0)$ cm.

Length of the third semicircle ($l_3$) with radius $r_3 = 1.5$ cm $= \pi r_3 = \pi(1.5)$ cm.

Length of the fourth semicircle ($l_4$) with radius $r_4 = 2.0$ cm $= \pi r_4 = \pi(2.0)$ cm.

And so on.

---

The lengths of the successive semicircles form a sequence:

$0.5\pi, 1.0\pi, 1.5\pi, 2.0\pi, \dots$

---

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

$l_2 - l_1 = 1.0\pi - 0.5\pi = 0.5\pi$

$l_3 - l_2 = 1.5\pi - 1.0\pi = 0.5\pi$

$l_4 - l_3 = 2.0\pi - 1.5\pi = 0.5\pi$

Since the difference between consecutive terms is constant, the sequence of lengths forms an AP.

---

In this AP of lengths:

The first term is $a = l_1 = 0.5\pi$.

The common difference is $d = 0.5\pi$.

The spiral is made up of thirteen consecutive semicircles, so the number of terms is $n=13$.

---

We need to find the total length of the spiral, which is the sum of the lengths of the thirteen semicircles. This is the sum of the first 13 terms of this AP ($S_{13}$).

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

---

Substitute the values $a = 0.5\pi$, $d = 0.5\pi$, and $n = 13$ into formula (1) to find the total length ($S_{13}$):

$S_{13} = \frac{13}{2}[2(0.5\pi) + (13-1)(0.5\pi)]$

$S_{13} = \frac{13}{2}[\pi + (12)(0.5\pi)]$

$S_{13} = \frac{13}{2}[\pi + 6\pi]$

$S_{13} = \frac{13}{2}[7\pi]$

$S_{13} = \frac{13 \times 7\pi}{2}$

---

Now substitute the value of $\pi = \frac{22}{7}$ into equation (i):

$S_{13} = \frac{91}{2} \times \frac{22}{7}$

$S_{13} = \frac{\cancel{91}^{13}}{2} \times \frac{22}{\cancel{7}_1}$

$S_{13} = \frac{13}{2} \times 22$

$S_{13} = 13 \times \frac{22}{2}$

$S_{13} = 13 \times 11$

$S_{13} = 143$.

---

Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

---

Conclusion: The total length of the spiral is $\mathbf{143}$ cm.

The number of logs in each row, starting from the bottom, forms a sequence:

Bottom row: 20 logs

Next row: 19 logs

Row next to that: 18 logs

and so on.

---

The sequence of the number of logs in the rows is $20, 19, 18, \dots$.

This sequence forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 20$.

The common difference is $d = 19 - 20 = -1$.

The total number of logs is 200. This is the sum of the logs in all the rows. Let the number of rows be $n$. The sum of the first $n$ terms ($S_n$) is 200.

$S_n = 200$.

---

We need to find the number of rows ($n$) and the number of logs in the top row (which is the $n$-th term, $a_n$).

---

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

---

Substitute the values $S_n = 200$, $a = 20$, and $d = -1$ into formula (1):

$200 = \frac{n}{2}[2(20) + (n-1)(-1)]$

$200 = \frac{n}{2}[40 - (n-1)]$

$200 = \frac{n}{2}[40 - n + 1]$

$200 = \frac{n}{2}[41 - n]$

Multiply both sides by 2:

$2 \times 200 = n(41 - n)$

$400 = 41n - n^2$

Rearrange the terms to form a quadratic equation:

$n^2 - 41n + 400 = 0$

---

Now we solve this quadratic equation for $n$. We can use factorization or the quadratic formula. We look for two numbers that multiply to 400 and add up to -41. These numbers are -16 and -25.

$n^2 - 16n - 25n + 400 = 0$

$n(n - 16) - 25(n - 16) = 0$

$(n - 16)(n - 25) = 0$

---

This gives two possible values for $n$:

$n - 16 = 0 \implies n = 16$

$n - 25 = 0 \implies n = 25$

---

Both $n=16$ and $n=25$ are positive integers. We need to check if both are valid solutions in the context of the problem.

The number of logs in each successive row is decreasing. Let's find the number of logs in the top row ($a_n$) for each value of $n$ using the formula $a_n = a + (n-1)d$, with $a=20$ and $d=-1$.

If $n=16$:

$a_{16} = 20 + (16-1)(-1)$

$a_{16} = 20 + (15)(-1)$

$a_{16} = 20 - 15 = 5$.

A row having 5 logs is possible.

If $n=25$:

$a_{25} = 20 + (25-1)(-1)$

$a_{25} = 20 + (24)(-1)$

$a_{25} = 20 - 24 = -4$.

The number of logs in a row cannot be negative. Therefore, $n=25$ is not a valid solution in this context.

Thus, the number of rows is 16.

---

The number of rows is $n=16$.

The number of logs in the top row is the 16th term, $a_{16}$, which we calculated as 5.

---

Conclusion: The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

There are 10 potatoes in the line.

Distance of the first potato from the bucket = 5 m.

Distance between consecutive potatoes = 3 m.

---

To pick up the first potato:

Distance run to the first potato = 5 m.

Distance run back to the bucket = 5 m.

Total distance for the first potato = $5 + 5 = 2 \times 5 = 10$ m.

---

To pick up the second potato:

The second potato is at a distance of $5 + 3 = 8$ m from the bucket.

Distance run to the second potato = 8 m.

Distance run back to the bucket = 8 m.

Total distance for the second potato = $8 + 8 = 2 \times 8 = 16$ m.

---

To pick up the third potato:

The third potato is at a distance of $5 + 3 + 3 = 5 + 2 \times 3 = 11$ m from the bucket.

Distance run to the third potato = 11 m.

Distance run back to the bucket = 11 m.

Total distance for the third potato = $11 + 11 = 2 \times 11 = 22$ m.

---

The distances run to pick up each successive potato and return to the bucket form a sequence:

Distance for 1st potato: $2 \times 5 = 10$ m

Distance for 2nd potato: $2 \times (5+3) = 2 \times 8 = 16$ m

Distance for 3rd potato: $2 \times (5+3+3) = 2 \times 11 = 22$ m

Distance for 4th potato: $2 \times (5+3+3+3) = 2 \times 14 = 28$ m

and so on.

---

The sequence of distances run for each potato is $10, 16, 22, 28, \dots$.

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

$16 - 10 = 6$

$22 - 16 = 6$

$28 - 22 = 6$

Since the difference is constant, the sequence forms an AP with common difference $d = 6$.

---

In this AP:

The first term is $a = 10$.

The common difference is $d = 6$.

There are 10 potatoes, so we need to find the total distance run to pick up all 10 potatoes. This is the sum of the first 10 terms of this AP ($S_{10}$).

The number of terms is $n=10$.

---

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

---

Substitute the values $a = 10$, $d = 6$, and $n = 10$ into formula (1) to find the total distance ($S_{10}$):

$S_{10} = \frac{10}{2}[2(10) + (10-1)6]$

$S_{10} = 5[20 + (9)6]$

$S_{10} = 5[20 + 54]$

$S_{10} = 5[74]$

Calculate $5 \times 74$:

$5 \times 74 = 5 \times (70 + 4) = 350 + 20 = 370$.

$S_{10} = 370$.

---

Thus, the total distance the competitor has to run is 370 meters.

---

Conclusion: The total distance the competitor has to run is $\mathbf{370}$ m.

The given Arithmetic Progression (AP) is: $121, 117, 113, \dots$

---

Here, the first term is $a = 121$.

The common difference $d$ is the difference between consecutive terms:

$d = 117 - 121 = -4$.

We can check this with the next pair: $113 - 117 = -4$. So, $d = -4$.

---

We are looking for the first negative term. Let the $n$-th term ($a_n$) be the first term that is negative. This means we need to find the smallest positive integer $n$ such that $a_n < 0$.

---

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

---

We set $a_n < 0$ and substitute the values of $a$ and $d$ into formula (1):

$a + (n-1)d < 0$

$121 + (n-1)(-4) < 0$

$121 - 4(n-1) < 0$

Subtract 121 from both sides:

$-4(n-1) < -121$

Divide both sides by -4. When dividing an inequality by a negative number, the inequality sign reverses.

$n-1 > \frac{121}{4}$

Convert the fraction to a decimal or mixed number:

$\frac{121}{4} = \frac{120+1}{4} = \frac{120}{4} + \frac{1}{4} = 30 + 0.25 = 30.25$.

So, the inequality is:

$n-1 > 30.25$

Add 1 to both sides:

$n > 30.25 + 1$

$n > 31.25$

---

Since $n$ must be an integer (representing the term number), the smallest integer greater than 31.25 is 32.

So, the first value of $n$ for which $a_n$ is negative is $n = 32$.

---

Let's check the 31st and 32nd terms to confirm:

$a_{31} = 121 + (31-1)(-4) = 121 + 30(-4) = 121 - 120 = 1$ (positive).

$a_{32} = 121 + (32-1)(-4) = 121 + 31(-4) = 121 - 124 = -3$ (negative).

This confirms that the 32nd term is the first negative term.

---

Conclusion: The 32nd term of the AP is its first negative term.

Given:

In an Arithmetic Progression (AP):

Sum of the third and seventh terms is 6.

Product of the third and seventh terms 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 $a_n = a + (n-1)d$.

Therefore, the third term ($a_3$) is:

$a_3 = a + (3-1)d = a + 2d$

And the seventh term ($a_7$) is:

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

According to the first condition, their sum is 6:

$a_3 + a_7 = 6$

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

$2a + 8d = 6$

Dividing the equation by 2, we get:

From this, we can express $a$ in terms of $d$:

According to the second condition, their product is 8:

$a_3 \times a_7 = 8$

Substitute the value of $a$ from equation (i) into equation (ii):

$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$

$(3 - 2d)(3 + 2d) = 8$

Using the algebraic identity $(x-y)(x+y) = x^2 - y^2$:

$3^2 - (2d)^2 = 8$

$9 - 4d^2 = 8$

$4d^2 = 9 - 8$

$4d^2 = 1$

$d^2 = \frac{1}{4}$

$d = \pm\sqrt{\frac{1}{4}}$

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

We have two possible values for the common difference $d$. This means there are two possible APs.

Case 1: When $d = \frac{1}{2}$

Substitute $d = \frac{1}{2}$ into equation (i):

$a = 3 - 4\left(\frac{1}{2}\right) = 3 - 2 = 1$

So, for this AP, $a = 1$ and $d = \frac{1}{2}$.

The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.

We need to find the sum of the first sixteen terms ($S_{16}$):

$S_{16} = \frac{16}{2}\left[2(1) + (16-1)\left(\frac{1}{2}\right)\right]$

$S_{16} = 8\left[2 + 15\left(\frac{1}{2}\right)\right]$

$S_{16} = 8\left[2 + \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{4}{2} + \frac{15}{2}\right]$

$S_{16} = 8\left(\frac{19}{2}\right)$

$S_{16} = 4 \times 19 = 76$

Case 2: When $d = -\frac{1}{2}$

Substitute $d = -\frac{1}{2}$ into equation (i):

$a = 3 - 4\left(-\frac{1}{2}\right) = 3 + 2 = 5$

So, for this AP, $a = 5$ and $d = -\frac{1}{2}$.

Now, we find the sum of the first sixteen terms ($S_{16}$):

$S_{16} = \frac{16}{2}\left[2(5) + (16-1)\left(-\frac{1}{2}\right)\right]$

$S_{16} = 8\left[10 + 15\left(-\frac{1}{2}\right)\right]$

$S_{16} = 8\left[10 - \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{20}{2} - \frac{15}{2}\right]$

$S_{16} = 8\left(\frac{5}{2}\right)$

$S_{16} = 4 \times 5 = 20$

---

Hence, there are two possible sums for the first sixteen terms of the AP. The sum is either 20 or 76.

The distance between consecutive rungs is 25 cm.

The distance between the top and bottom rungs is $2\frac{1}{2}$ m.

Convert the distance between top and bottom rungs to centimeters: $2\frac{1}{2}$ m $= 2.5$ m $= 2.5 \times 100$ cm $= 250$ cm.

---

Let the number of rungs be $n$. The distance between $n$ rungs is $(n-1)$ times the distance between consecutive rungs.

Distance between top and bottom rungs = $(n-1) \times$ Distance between consecutive rungs.

$250 = (n-1) \times 25$

Divide both sides by 25:

$\frac{250}{25} = n-1$

$10 = n-1$

$n = 10 + 1 = 11$.

So, there are 11 rungs on the ladder.

---

The lengths of the rungs decrease uniformly from bottom to top. This means the lengths form an Arithmetic Progression (AP).

The length of the bottom rung is 45 cm. This is the first term, $a = 45$.

The length of the top rung is 25 cm. This is the last term, $a_n$. Since there are 11 rungs, the top rung is the 11th term, $a_{11} = 25$.

The number of terms is $n = 11$.

---

We need to find the total length of the wood required for the rungs, which is the sum of the lengths of all 11 rungs. This is the sum of the first 11 terms of this AP ($S_{11}$).

We can use the formula for the sum of the first $n$ terms when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

---

Substitute the values $n=11$, $a=45$, and $a_{11}=25$ into formula (1):

$S_{11} = \frac{11}{2}(45 + 25)$

$S_{11} = \frac{11}{2}(70)$

$S_{11} = 11 \times \frac{70}{2}$

$S_{11} = 11 \times 35$

$S_{11} = 385$.

---

Thus, the total length of the wood required for the rungs is 385 cm.

---

Conclusion: The length of the wood required for the rungs is $\mathbf{385}$ cm.

The house numbers are $1, 2, 3, \dots, 49$. This is an Arithmetic Progression (AP) with the first term $a=1$ and the common difference $d=1$.

---

The sum of the numbers of the houses preceding the house numbered $x$ is the sum of the first $x-1$ house numbers. This is the sum of the first $x-1$ terms of the AP, $S_{x-1}$.

For $x=1$, there are no houses preceding it, so the sum is 0. $S_{1-1} = S_0 = 0$.

For $x > 1$, $S_{x-1} = 1 + 2 + \dots + (x-1)$.

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$ with $n=x-1$, $a=1$, and $a_{x-1}=x-1$:

---

The sum of the numbers of the houses following the house numbered $x$ is the sum of the house numbers from $x+1$ to 49. This sum can be found by subtracting the sum of the first $x$ house numbers ($S_x$) from the total sum of all 49 house numbers ($S_{49}$).

The houses following house $x$ are $x+1, x+2, \dots, 49$. This is also an AP with first term $a' = x+1$, last term $l'=49$, and number of terms $n' = 49 - (x+1) + 1 = 49 - x$. The sum is $\frac{49-x}{2}((x+1)+49) = \frac{49-x}{2}(x+50)$.

Alternatively, using the hint, the sum of houses following $x$ is $S_{49} - S_x$.

The total sum of houses from 1 to 49 ($S_{49}$) is the sum of an AP with $a=1$, $n=49$, and $a_{49}=49$.

$S_{49} = \frac{49}{2}(1 + 49) = \frac{49}{2}(50) = 49 \times 25$.

$49 \times 25 = (50-1) \times 25 = 1250 - 25 = 1225$.

The sum of the first $x$ house numbers ($S_x$) is the sum of an AP with $a=1$, $n=x$, and $a_x=x$.

The sum of the numbers of the houses following house $x$ is $S_{49} - S_x$.

---

We are given that the sum of the numbers of the houses preceding house $x$ is equal to the sum of the numbers of the houses following it.

Substitute the expressions from (i) and (iv) into this equation:

$\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$

Add $\frac{x(x+1)}{2}$ to both sides:

$\frac{x(x-1)}{2} + \frac{x(x+1)}{2} = 1225$

Combine the terms on the left side:

$\frac{x(x-1) + x(x+1)}{2} = 1225$

$\frac{x^2 - x + x^2 + x}{2} = 1225$

$\frac{2x^2}{2} = 1225$

$x^2 = 1225$

Take the square root of both sides:

$x = \pm \sqrt{1225}$.

To find $\sqrt{1225}$:

$\begin{array}{c|cc} & 3 \ 5 & \\ \hline \phantom{()} 3 & \overline{12} \; \overline{25} & \\ + \; 3 & 9\phantom{........)} \\ \hline \phantom{()} 6 \; 5 & 3 \; 25 \phantom{)} \\ \phantom{()} +5 & 3 \; 25 \phantom{)} \\ \hline \phantom{()} 70 & 0\phantom{)} \end{array}$

So, $\sqrt{1225} = 35$.

$x = \pm 35$.

---

Since $x$ represents a house number, it must be a positive integer. Also, the house numbered $x$ must be within the range of 1 to 49. Thus, $x=35$ is the only valid solution.

The value of $x=35$ lies between 1 and 49 (exclusive of 1, since $S_{x-1}$ requires $x-1 \geq 0$). For $x=35$, $x-1=34$, and houses following 35 exist (up to 49).

---

Conclusion: There is a value of $x$ that satisfies the condition, and that value is $\mathbf{35}$.

Given:

A terrace at a football ground has 15 steps.

For each step:

• Length = 50 m
• Rise (height) = $\frac{1}{4}$ m
• Tread (width) = $\frac{1}{2}$ m

---

To Find:

The total volume of concrete required to build the terrace.

---

Solution:

The terrace is built step-by-step, where each subsequent step is built on top of the previous one. This means the height of the concrete structure increases with each step from the front.

Let's calculate the volume of concrete required for each step, starting from the first (bottom) step.

The volume of a step can be calculated as a cuboid: Volume = Length × Tread × Height.

Volume of concrete for the 1st step:

The height of the first step from the ground is $1 \times (\text{rise}) = 1 \times \frac{1}{4} = \frac{1}{4}$ m.

$V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{50}{8} = \frac{25}{4}$ m³.

Volume of concrete for the 2nd step:

The height of the second step from the ground is $2 \times (\text{rise}) = 2 \times \frac{1}{4} = \frac{2}{4}$ m.

$V_2 = 50 \times \frac{1}{2} \times \frac{2}{4} = \frac{100}{8} = \frac{50}{4}$ m³.

Volume of concrete for the 3rd step:

The height of the third step from the ground is $3 \times (\text{rise}) = 3 \times \frac{1}{4} = \frac{3}{4}$ m.

$V_3 = 50 \times \frac{1}{2} \times \frac{3}{4} = \frac{150}{8} = \frac{75}{4}$ m³.

We can see that the volumes of concrete required for each step form an Arithmetic Progression (AP):

$\frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \dots$

Here, the first term is $a = \frac{25}{4}$.

The common difference is $d = V_2 - V_1 = \frac{50}{4} - \frac{25}{4} = \frac{25}{4}$.

The number of terms (steps) is $n = 15$.

The total volume of concrete is the sum of the volumes of all 15 steps, which is the sum of the first 15 terms of this AP ($S_{15}$).

The formula for the sum of an AP is:

Substituting the values:

$S_{15} = \frac{15}{2}\left[2\left(\frac{25}{4}\right) + (15-1)\left(\frac{25}{4}\right)\right]$

$S_{15} = \frac{15}{2}\left[\frac{50}{4} + 14\left(\frac{25}{4}\right)\right]$

$S_{15} = \frac{15}{2}\left[\frac{50}{4} + \frac{350}{4}\right]$

$S_{15} = \frac{15}{2}\left[\frac{50 + 350}{4}\right]$

$S_{15} = \frac{15}{2}\left[\frac{400}{4}\right]$

$S_{15} = \frac{15}{2} \times 100$

$S_{15} = 15 \times 50$

$S_{15} = 750$

Thus, the total volume of concrete required is 750 m³.

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Alternate Solution:

The total volume can be calculated by factoring out the constant dimensions and summing the changing heights.

Total Volume = (Volume of 1st step) + (Volume of 2nd step) + ... + (Volume of 15th step)

Total Volume = $\left(50 \times \frac{1}{2} \times \frac{1}{4}\right) + \left(50 \times \frac{1}{2} \times \frac{2}{4}\right) + \dots + \left(50 \times \frac{1}{2} \times \frac{15}{4}\right)$

We can factor out the common term $50 \times \frac{1}{2} \times \frac{1}{4}$:

Total Volume = $50 \times \frac{1}{2} \times \frac{1}{4} \times (1 + 2 + 3 + \dots + 15)$

The expression in the parentheses is the sum of the first 15 natural numbers. The formula for the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

Sum of first 15 natural numbers = $\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$.

Now, substitute this back into the volume equation:

Total Volume = $\left(50 \times \frac{1}{2} \times \frac{1}{4}\right) \times 120$

Total Volume = $\frac{50}{8} \times 120$

Total Volume = $\frac{25}{4} \times 120$

Total Volume = $25 \times 30 = 750$ m³.

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Hence, the total volume of concrete required to build the terrace is 750 m³.