NCERT Solutions For Class 10 Maths: Arithmetic Progression

EXERCISE 5.1

Question 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Solution-

The list of numbers representing the taxi fare (in ₹) is 15, $15 + 8$, $15 + 8 + 8$, and so on.

The sequence is $15, 23, 31, 39, \dots$.

The difference between successive terms ($a_{k+1} – a_k$) is $23 - 15 = 8$, $31 - 23 = 8$, etc.. Since the difference is a fixed number ($d=8$) added to the preceding term, the list of numbers forms an arithmetic progression (AP).

(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.

Solution-

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

After the first pump removal, remaining air: $V - \frac{1}{4}V = \frac{3}{4}V$.

After the second removal, remaining air: $\frac{3}{4}V - \frac{1}{4}\left(\frac{3}{4}V\right) = \frac{3}{4}V \left(1 - \frac{1}{4}\right) = \left(\frac{3}{4}\right)^2 V$.

The list of numbers is $V, \frac{3}{4}V, \left(\frac{3}{4}\right)^2 V, \dots$.

The difference between the first two terms is $d_1 = \frac{3}{4}V - V = -\frac{1}{4}V$.

The difference between the next two terms is $d_2 = \left(\frac{3}{4}\right)^2 V - \frac{3}{4}V = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$.

Since $d_1 \neq d_2$, the difference between consecutive terms is not the same. Therefore, the list of numbers does not form an AP.

(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

Solution-

The cost (in ₹) for $1 \text{ m}$ is 150.

The cost for $2 \text{ m}$ is $150 + 50 = 200$.

The cost for $3 \text{ m}$ is $200 + 50 = 250$.

The list of numbers is $150, 200, 250, 300, \dots$.

The fixed number added to the preceding term is ₹ 50.

Therefore, the list of numbers forms an arithmetic progression (AP) with a common difference $d=50$.

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.

Solution-

Initial amount $= 10000$.

Amount after 1 year $= 10000 \left(1 + \frac{8}{100}\right) = 10000(1.08)$.

Amount after 2 years $= 10000(1.08)^2$.

The list of numbers is $10000, 10000(1.08), 10000(1.08)^2, \dots$.

The differences are: $d_1 = 10000(1.08) - 10000 = 800$.

$d_2 = 10000(1.08)^2 - 10000(1.08) = 10000(1.08)(0.08) = 864$.

Since $d_1 \neq d_2$, the difference between consecutive terms is not the same. Therefore, the list of numbers does not form an AP.

Question 2: Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:

(i) a = 10, d = 10

Solution-

The general form of an AP is $a, a + d, a + 2d, a + 3d, \dots$.

First term $a_1 = a = 10$.

Second term $a_2 = 10 + 10 = 20$.

Third term $a_3 = 20 + 10 = 30$.

Fourth term $a_4 = 30 + 10 = 40$.

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

(ii) a = –2, d = 0

Solution-

First term $a_1 = a = -2$.

Second term $a_2 = -2 + 0 = -2$.

Third term $a_3 = -2 + 0 = -2$.

Fourth term $a_4 = -2 + 0 = -2$.

The first four terms are –2, –2, –2, –2.

(iii) a = 4, d = – 3

Solution-

First term $a_1 = a = 4$.

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

Third term $a_3 = 1 + (-3) = -2$.

Fourth term $a_4 = -2 + (-3) = -5$.

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

(iv) a = – 1, d = $\frac{1}{2}$

Solution-

First term $a_1 = a = -1$.

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

Third term $a_3 = -\frac{1}{2} + \frac{1}{2} = 0$.

Fourth term $a_4 = 0 + \frac{1}{2} = \frac{1}{2}$.

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

(v) a = – 1.25, d = – 0.25

Solution-

First term $a_1 = a = -1.25$.

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

Third term $a_3 = -1.50 + (-0.25) = -1.75$.

Fourth term $a_4 = -1.75 + (-0.25) = -2.00$.

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

Question 3: For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .

Solution-

First term $a$ is the initial term in the list.
$a = 3$.

Common difference $d$ is found by subtracting any term from its succeeding term ($a_{k+1} - a_k$).
$d = 1 - 3 = -2$.
(Check: $-1 - 1 = -2$; $-3 - (-1) = -2$).

First term $a$ is 3 and common difference $d$ is –2.

(ii) – 5, – 1, 3, 7, . . .

Solution-

First term $a = -5$.

Common difference $d = -1 - (-5) = -1 + 5 = 4$.
(Check: $3 - (-1) = 4$; $7 - 3 = 4$).

First term $a$ is –5 and common difference $d$ is 4.

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

Solution-

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

Common difference $d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$.
(Check: $\frac{9}{3} - \frac{5}{3} = \frac{4}{3}$).

First term $a$ is $\mathbf{\frac{1}{3}}$ and common difference $d$ is $\mathbf{\frac{4}{3}}$.

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

Solution-

First term $a = 0.6$.

Common difference $d = 1.7 - 0.6 = 1.1$.
(Check: $2.8 - 1.7 = 1.1$).

First term $a$ is 0.6 and common difference $d$ is 1.1.

Question 4: Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms.

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

Solution-

Difference $a_2 - a_1 = 4 - 2 = 2$.

Difference $a_3 - a_2 = 8 - 4 = 4$.

Since $a_2 - a_1 \neq a_3 - a_2$, the difference between consecutive terms is not the same.

The list does not form an AP.

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

Solution-

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

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

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

Since $a_{k+1} - a_k$ is the same every time ($d = \frac{1}{2}$), the list forms an AP.

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

The next three terms are:
$a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$.

$a_6 = 4 + \frac{1}{2} = \frac{9}{2}$.

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

The common difference is $\mathbf{\frac{1}{2}}$ and the next three terms are $\mathbf{4, \frac{9}{2}, 5}$.

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

Solution-

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

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

Since $d = -2.0$ is fixed, the list forms an AP.

Common difference $d = -2.0$.

The next three terms are:
$a_5 = -7.2 + (-2.0) = -9.2$.

$a_6 = -9.2 + (-2.0) = -11.2$.

$a_7 = -11.2 + (-2.0) = -13.2$.

The common difference is –2.0 and the next three terms are –9.2, –11.2, –13.2.

(iv) – 10, – 6, – 2, 2, . . .

Solution-

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

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

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

Since $d=4$ is fixed, the list forms an AP.

Common difference $d = 4$.

The next three terms are:
$a_5 = 2 + 4 = 6$.

$a_6 = 6 + 4 = 10$.

$a_7 = 10 + 4 = 14$.

The common difference is 4 and the next three terms are 6, 10, 14.

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

Solution-

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

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

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

Since $d = \sqrt{2}$ is fixed, the list forms an AP.

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

The next three terms are:
$a_5 = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$.

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

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

The common difference is $\mathbf{\sqrt{2}}$ and the next three terms are $\mathbf{3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}}$.

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

Solution-

Difference $a_2 - a_1 = 0.22 - 0.2 = 0.02$.

Difference $a_3 - a_2 = 0.222 - 0.22 = 0.002$.

Since $a_2 - a_1 \neq a_3 - a_2$, the difference between consecutive terms is not the same.

The list does not form an AP.

(vii) 0, – 4, – 8, –12, . . .

Solution-

Difference $a_2 - a_1 = -4 - 0 = -4$.

Difference $a_3 - a_2 = -8 - (-4) = -4$.

Difference $a_4 - a_3 = -12 - (-8) = -4$.

Since $d = -4$ is fixed, the list forms an AP.

Common difference $d = -4$.

The next three terms are:
$a_5 = -12 + (-4) = -16$.

$a_6 = -16 + (-4) = -20$.

$a_7 = -20 + (-4) = -24$.

The common difference is –4 and the next three terms are –16, –20, –24.

(viii) – $\frac{1}{2}$, – $\frac{1}{2}$, – $\frac{1}{2}$, – $\frac{1}{2}$, . . .

Solution-

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

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

Since $d = 0$ is fixed (d can be positive, negative, or zero), the list forms an AP.

Common difference $d = 0$.

The next three terms are:
$a_5 = -\frac{1}{2} + 0 = -\frac{1}{2}$.

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

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

The common difference is 0 and the next three terms are $\mathbf{-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}}$.

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

Solution-

Difference $a_2 - a_1 = 3 - 1 = 2$.

Difference $a_3 - a_2 = 9 - 3 = 6$.

Since $a_2 - a_1 \neq a_3 - a_2$, the difference between consecutive terms is not the same.

The list does not form an AP.

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

Solution-

Difference $a_2 - a_1 = 2a - a = a$.

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

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

Since $d = a$ is fixed, the list forms an AP.

Common difference $d = a$.

The next three terms are:
$a_5 = 4a + a = 5a$.

$a_6 = 5a + a = 6a$.

$a_7 = 6a + a = 7a$.

The common difference is $\mathbf{a}$ and the next three terms are $\mathbf{5a, 6a, 7a}$.

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

Solution-

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

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

Since $a(a-1) \neq a^2(a-1)$ (unless $a=1$ or $a=0$), the difference is not constant.

The list does not form an AP.

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

Solution-

Rewrite the terms: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$

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

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

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

Since $d = \sqrt{2}$ is fixed, the list forms an AP.

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

The next three terms are:
$a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}$.

$a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$.

$a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$.

The common difference is $\mathbf{\sqrt{2}}$ and the next three terms are $\mathbf{\sqrt{50}, \sqrt{72}, \sqrt{98}}$.

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

Solution-

Rewrite terms: $\sqrt{3}, \sqrt{6}, 3, 2\sqrt{3}, \dots$

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

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

Since $\sqrt{6} - \sqrt{3} \neq 3 - \sqrt{6}$, the difference is not constant.

The list does not form an AP.

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

Solution-

The list of numbers is $1, 9, 25, 49, \dots$

Difference $a_2 - a_1 = 9 - 1 = 8$.

Difference $a_3 - a_2 = 25 - 9 = 16$.

Since $a_2 - a_1 \neq a_3 - a_2$, the difference between consecutive terms is not the same.

The list does not form an AP.

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

Solution-

The list of numbers is $1, 25, 49, 73, \dots$

Difference $a_2 - a_1 = 25 - 1 = 24$.

Difference $a_3 - a_2 = 49 - 25 = 24$.

Difference $a_4 - a_3 = 73 - 49 = 24$.

Since $d = 24$ is fixed, the list forms an AP.

Common difference $d = 24$.

The next three terms are:
$a_5 = 73 + 24 = 97$.

$a_6 = 97 + 24 = 121$.

$a_7 = 121 + 24 = 145$.

The common difference is 24 and the next three terms are 97, 121, 145.

EXERCISE 5.2

Question 1: Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP:

Solution-

The formula for the $n$th term is $a_n = a + (n – 1) d$.

(i) Given $a = 7, d = 3, n = 8$. Find $a_n$.
$a_8 = 7 + (8 - 1) \times 3$
$a_8 = 7 + 7 \times 3 = 7 + 21 = 28$.

$a_n = \mathbf{28}$.

(ii) Given $a = – 18, n = 10, a_n = 0$. Find $d$.
$0 = -18 + (10 - 1) d$
$0 = -18 + 9d$
$9d = 18$
$d = 2$.

$d = \mathbf{2}$.

(iii) Given $d = – 3, n = 18, a_n = – 5$. Find $a$.
$-5 = a + (18 - 1) (-3)$
$-5 = a + 17 \times (-3)$
$-5 = a - 51$
$a = 51 - 5 = 46$.

$a = \mathbf{46}$.

(iv) Given $a = – 18.9, d = 2.5, a_n = 3.6$. Find $n$.
$3.6 = -18.9 + (n - 1) (2.5)$
$3.6 + 18.9 = (n - 1) (2.5)$
$22.5 = (n - 1) (2.5)$
$n - 1 = \frac{22.5}{2.5} = 9$
$n = 10$.

$n = \mathbf{10}$.

(v) Given $a = 3.5, d = 0, n = 105$. Find $a_n$.
$a_{105} = 3.5 + (105 - 1) \times 0$
$a_{105} = 3.5 + 0 = 3.5$.

$a_n = \mathbf{3.5}$.

Question 2: Choose the correct choice in the following and justify:

(i) 30th term of the AP: $10, 7, 4, \dots$, is

(A) 97 (B) 77 (C) –77 (D) – 87

Solution-

Here, $a = 10$, $d = 7 - 10 = -3$, and $n = 30$.
Using $a_n = a + (n – 1) d$:
$a_{30} = 10 + (30 - 1) (-3)$
$a_{30} = 10 + 29 \times (-3)$
$a_{30} = 10 - 87$
$a_{30} = -77$.

The correct choice is (C) –77.

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

(A) 28 (B) 22 (C) –38 (D) – $48\frac{1}{2}$

Solution-

Here, $a = -3$, $n = 11$.
$d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{5}{2}$.
Using $a_n = a + (n – 1) d$:
$a_{11} = -3 + (11 - 1) \times \frac{5}{2}$
$a_{11} = -3 + 10 \times \frac{5}{2}$
$a_{11} = -3 + 25$
$a_{11} = 22$.

The correct choice is (B) 22.

Question 3: In the following APs, find the missing terms in the boxes:

(i) 2, $\mathbf{\square}$, 26

Solution-

Given $a_1 = 2$, $a_3 = 26$. We need $a_2$.
$a_3 = a_1 + 2d$.
$26 = 2 + 2d$
$24 = 2d$
$d = 12$.
$a_2 = a_1 + d = 2 + 12 = 14$.

The missing term is 14.

(ii) $\mathbf{\square}$, 13, $\mathbf{\square}$, 3

Solution-

Given $a_2 = 13$, $a_4 = 3$.
$a_2 = a + d = 13$ (1)
$a_4 = a + 3d = 3$ (2)
Subtracting (1) from (2):
$(a + 3d) - (a + d) = 3 - 13$
$2d = -10$
$d = -5$.
Substitute $d=-5$ into (1):
$a + (-5) = 13$
$a = 18$.
$a_3 = a + 2d = 18 + 2(-5) = 18 - 10 = 8$.

The missing terms are 18 and 8.

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

Solution-

Given $a_1 = 5$, $a_4 = 9\frac{1}{2} = \frac{19}{2}$.
$a_4 = a + 3d$.
$\frac{19}{2} = 5 + 3d$
$3d = \frac{19}{2} - 5 = \frac{19 - 10}{2} = \frac{9}{2}$
$d = \frac{9}{2} \times \frac{1}{3} = \frac{3}{2}$.
$a_2 = a + d = 5 + \frac{3}{2} = \frac{10 + 3}{2} = \frac{13}{2} = 6\frac{1}{2}$.
$a_3 = a + 2d = 5 + 2\left(\frac{3}{2}\right) = 5 + 3 = 8$.

The missing terms are $\mathbf{6\frac{1}{2}}$ and 8.

(iv) – 4, $\mathbf{\square}$, $\mathbf{\square}$, $\mathbf{\square}$, $\mathbf{\square}$, 6

Solution-

Given $a_1 = -4$, $a_6 = 6$.
$a_6 = a + 5d$.
$6 = -4 + 5d$
$10 = 5d$
$d = 2$.
$a_2 = -4 + 2 = -2$.
$a_3 = -2 + 2 = 0$.
$a_4 = 0 + 2 = 2$.
$a_5 = 2 + 2 = 4$.

The missing terms are –2, 0, 2, 4.

(v) $\mathbf{\square}$, 38, $\mathbf{\square}$, $\mathbf{\square}$, $\mathbf{\square}$, – 22

Solution-

Given $a_2 = 38$, $a_6 = -22$.
$a_2 = a + d = 38$ (1)
$a_6 = a + 5d = -22$ (2)
Subtracting (1) from (2):
$4d = -22 - 38 = -60$
$d = -15$.
Substitute $d=-15$ into (1):
$a + (-15) = 38$
$a = 53$.
$a_3 = 38 + (-15) = 23$.
$a_4 = 23 + (-15) = 8$.
$a_5 = 8 + (-15) = -7$.

The missing terms are 53, 23, 8, –7.

Question 4: Which term of the AP : 3, 8, 13, 18, . . . , is 78?

Solution-

Given $a = 3$, $d = 8 - 3 = 5$. Let $a_n = 78$. We need to find $n$.
Using $a_n = a + (n – 1) d$:
$78 = 3 + (n - 1) \times 5$
$75 = (n - 1) \times 5$
$n - 1 = \frac{75}{5} = 15$
$n = 16$.

The 16th term of the AP is 78.

Question 5: Find the number of terms in each of the following APs:

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

Solution-

Given $a = 7$, $d = 13 - 7 = 6$, $a_n = l = 205$. We need to find $n$.
Using $a_n = a + (n – 1) d$:
$205 = 7 + (n - 1) \times 6$
$198 = (n - 1) \times 6$
$n - 1 = \frac{198}{6} = 33$
$n = 34$.

The number of terms is 34.

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

Solution-

Given $a = 18$, $a_n = l = -47$.
$d = 15 \frac{1}{2} - 18 = \frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2}$.
Using $a_n = a + (n – 1) d$:
$-47 = 18 + (n - 1) \left(-\frac{5}{2}\right)$
$-47 - 18 = (n - 1) \left(-\frac{5}{2}\right)$
$-65 = (n - 1) \left(-\frac{5}{2}\right)$
$n - 1 = -65 \times \left(-\frac{2}{5}\right) = 13 \times 2 = 26$
$n = 27$.

The number of terms is 27.

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Question 6: Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

Solution-

Given $a = 11$, $d = 8 - 11 = -3$. Let $a_n = -150$. We need to find $n$.
Using $a_n = a + (n – 1) d$:
$-150 = 11 + (n - 1) (-3)$
$-150 - 11 = -3(n - 1)$
$-161 = -3(n - 1)$
$n - 1 = \frac{161}{3}$
$n = \frac{161}{3} + 1 = \frac{164}{3}$.

Since $n$ must be a positive integer, and $\frac{164}{3}$ is not an integer, –150 is not a term of the given AP.

Question 7: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution-

Given $a_{11} = 38$ and $a_{16} = 73$.
Using $a_n = a + (n – 1) d$:
$a_{11} = a + 10d = 38$ (1)
$a_{16} = a + 15d = 73$ (2)
Subtracting (1) from (2):
$(a + 15d) - (a + 10d) = 73 - 38$
$5d = 35$
$d = 7$.
Substitute $d=7$ into (1):
$a + 10(7) = 38$
$a + 70 = 38$
$a = 38 - 70 = -32$.
Now find $a_{31}$:
$a_{31} = a + 30d$
$a_{31} = -32 + 30(7) = -32 + 210 = 178$.

The 31st term is 178.

Question 8: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution-

Total terms $n = 50$. Given $a_3 = 12$ and the last term $a_{50} = 106$.
$a_3 = a + 2d = 12$ (1)
$a_{50} = a + 49d = 106$ (2)
Subtracting (1) from (2):
$(a + 49d) - (a + 2d) = 106 - 12$
$47d = 94$
$d = 2$.
Substitute $d=2$ into (1):
$a + 2(2) = 12$
$a + 4 = 12$
$a = 8$.
Now find $a_{29}$:
$a_{29} = a + 28d$
$a_{29} = 8 + 28(2) = 8 + 56 = 64$.

The 29th term is 64.

Question 9: If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution-

Given $a_3 = 4$ and $a_9 = -8$.
$a_3 = a + 2d = 4$ (1)
$a_9 = a + 8d = -8$ (2)
Subtracting (1) from (2):
$6d = -8 - 4 = -12$
$d = -2$.
Substitute $d=-2$ into (1):
$a + 2(-2) = 4$
$a - 4 = 4$
$a = 8$.
Let $a_n = 0$. We need to find $n$.
$a_n = a + (n - 1) d$
$0 = 8 + (n - 1) (-2)$
$-8 = -2(n - 1)$
$4 = n - 1$
$n = 5$.

The 5th term of the AP is zero.

Question 10: The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution-

Given $a_{17} = a_{10} + 7$.
Using $a_n = a + (n – 1) d$:
$a_{17} = a + 16d$.
$a_{10} = a + 9d$.
$a + 16d = (a + 9d) + 7$
$16d - 9d = 7$
$7d = 7$
$d = 1$.

The common difference is 1.

Question 11: Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution-

Given $a = 3$, $d = 15 - 3 = 12$. Let $a_n$ be the required term.
We are given $a_n = a_{54} + 132$.
First find $a_{54}$:
$a_{54} = a + 53d = 3 + 53(12) = 3 + 636 = 639$.
$a_n = 639 + 132 = 771$.
Now find $n$ such that $a_n = 771$:
$a_n = a + (n - 1) d$
$771 = 3 + (n - 1) (12)$
$768 = 12(n - 1)$
$n - 1 = \frac{768}{12} = 64$
$n = 65$.

The 65th term will be 132 more than the 54th term.

Question 12: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution-

Let the first AP be denoted by $a, d$ and the second AP by $A, D$.
We are given $d = D$.
The difference between their 100th terms is 100:
$a_{100} - A_{100} = 100$
$(a + 99d) - (A + 99D) = 100$.
Since $d=D$:
$(a + 99d) - (A + 99d) = 100$
$a - A = 100$.
The difference between their 1000th terms is:
$a_{1000} - A_{1000} = (a + 999d) - (A + 999D)$.
Since $d=D$:
$a_{1000} - A_{1000} = (a + 999d) - (A + 999d) = a - A$.
Since $a - A = 100$, the difference between their 1000th terms is 100.

The difference between their 1000th terms is 100.

Question 13: How many three-digit numbers are divisible by 7?

Solution-

The list of three-digit numbers divisible by 7 forms an AP.
The first three-digit number divisible by 7 is 105.
The last three-digit number divisible by 7 is 994 ($999 = 7 \times 142 + 5$).
So, the AP is $105, 112, 119, \dots, 994$.
Here, $a = 105$, $d = 7$, $a_n = 994$. We need to find $n$.
Using $a_n = a + (n – 1) d$:
$994 = 105 + (n - 1) \times 7$
$889 = 7(n - 1)$
$n - 1 = \frac{889}{7} = 127$
$n = 128$.

There are 128 three-digit numbers divisible by 7.

Question 14: How many multiples of 4 lie between 10 and 250?

Solution-

The list of multiples of 4 between 10 and 250 forms an AP.
The first multiple of 4 greater than 10 is 12.
The last multiple of 4 less than 250 is 248 ($250 = 4 \times 62 + 2$).
So, the AP is $12, 16, 20, \dots, 248$.
Here, $a = 12$, $d = 4$, $a_n = 248$. We need to find $n$.
Using $a_n = a + (n – 1) d$:
$248 = 12 + (n - 1) \times 4$
$236 = 4(n - 1)$
$n - 1 = \frac{236}{4} = 59$
$n = 60$.

There are 60 multiples of 4 lying between 10 and 250.

Question 15: For what value of $n$, are the $n$th terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

Solution-

For the first AP: $a = 63$, $d = 65 - 63 = 2$.
The $n$th term is $a_n = 63 + (n - 1) 2 = 63 + 2n - 2 = 61 + 2n$.
For the second AP: $A = 3$, $D = 10 - 3 = 7$.
The $n$th term is $A_n = 3 + (n - 1) 7 = 3 + 7n - 7 = 7n - 4$.
Set $a_n = A_n$:
$61 + 2n = 7n - 4$
$61 + 4 = 7n - 2n$
$65 = 5n$
$n = 13$.

The $n$th terms are equal when $n = \mathbf{13}$.

Question 16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution-

Given $a_3 = 16$.
$a_3 = a + 2d = 16$ (1)
Given $a_7 = a_5 + 12$.
$a + 6d = (a + 4d) + 12$
$2d = 12$
$d = 6$.
Substitute $d=6$ into (1):
$a + 2(6) = 16$
$a + 12 = 16$
$a = 4$.
The AP is $a, a + d, a + 2d, \dots$.
The first three terms are 4, $4+6=10$, $10+6=16$.

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

Question 17: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution-

First term $a = 3$, common difference $d = 8 - 3 = 5$. Last term $l = 253$.
To find the 20th term from the last, we can reverse the AP.
In the reversed AP, the new first term $A = 253$, and the new common difference $D = -d = -5$.
We need the 20th term of the reversed AP, $A_{20}$:
$A_{20} = A + (20 - 1) D$
$A_{20} = 253 + 19 (-5)$
$A_{20} = 253 - 95 = 158$.

The 20th term from the last term is 158.

Question 18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution-

Given $a_4 + a_8 = 24$.
$(a + 3d) + (a + 7d) = 24$
$2a + 10d = 24$
$a + 5d = 12$ (1)
Given $a_6 + a_{10} = 44$.
$(a + 5d) + (a + 9d) = 44$
$2a + 14d = 44$
$a + 7d = 22$ (2)
Subtracting (1) from (2):
$(a + 7d) - (a + 5d) = 22 - 12$
$2d = 10$
$d = 5$.
Substitute $d=5$ into (1):
$a + 5(5) = 12$
$a + 25 = 12$
$a = -13$.
The first three terms are $a_1 = a = -13$, $a_2 = a+d = -13+5 = -8$, and $a_3 = a+2d = -8+5 = -3$.

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

Question 19: Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution-

The annual salary forms an AP with $a = 5000$ and $d = 200$. Let $a_n = 7000$.
$a_n = a + (n - 1) d$
$7000 = 5000 + (n - 1) 200$
$2000 = 200(n - 1)$
$n - 1 = \frac{2000}{200} = 10$
$n = 11$.
The 11th year corresponds to the salary of ₹ 7000.
Since he started in 1995 (1st year), the 11th year is $1995 + (11 - 1) = 2005$.

His income reached ₹ 7000 in the year 2005.

Question 20: Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$th week, her weekly savings become ₹ 20.75, find $n$.

Solution-

The weekly savings form an AP with $a = 5$ and $d = 1.75$. Let $a_n = 20.75$.
$a_n = a + (n - 1) d$
$20.75 = 5 + (n - 1) (1.75)$
$15.75 = 1.75(n - 1)$
$n - 1 = \frac{15.75}{1.75} = 9$
$n = 10$.

The value of $n$ is 10.

EXERCISE 5.3

Question 1: Find the sum of the following APs:

(i) $2, 7, 12, \dots$, to 10 terms.

Solution-

Given $a = 2$, $d = 7 - 2 = 5$, $n = 10$.
The sum formula is $S = \frac{n}{2} [2a + (n – 1) d]$.
$S_{10} = \frac{10}{2} [2(2) + (10 - 1) 5]$
$S_{10} = 5 [4 + 9 \times 5]$
$S_{10} = 5 [4 + 45] = 5 \times 49 = 245$.

The sum is 245.

(ii) –37, –33, –29, . . ., to 12 terms.

Solution-

Given $a = -37$, $d = -33 - (-37) = 4$, $n = 12$.
$S_{12} = \frac{12}{2} [2(-37) + (12 - 1) 4]$
$S_{12} = 6 [-74 + 11 \times 4]$
$S_{12} = 6 [-74 + 44] = 6 \times (-30) = -180$.

The sum is –180.

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.

Solution-

Given $a = 0.6$, $d = 1.7 - 0.6 = 1.1$, $n = 100$.
$S_{100} = \frac{100}{2} [2(0.6) + (100 - 1) 1.1]$
$S_{100} = 50 [1.2 + 99 \times 1.1]$
$S_{100} = 50 [1.2 + 108.9]$
$S_{100} = 50 [110.1] = 5505$.

The sum is 5505.

(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms.

Solution-

Given $a = \frac{1}{15}$, $n = 11$.
$d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$.
$S_{11} = \frac{11}{2} \left[2\left(\frac{1}{15}\right) + (11 - 1) \left(\frac{1}{60}\right)\right]$
$S_{11} = \frac{11}{2} \left[\frac{2}{15} + 10 \times \frac{1}{60}\right]$
$S_{11} = \frac{11}{2} \left[\frac{2}{15} + \frac{1}{6}\right]$
$S_{11} = \frac{11}{2} \left[\frac{4}{30} + \frac{5}{30}\right] = \frac{11}{2} \times \frac{9}{30}$
$S_{11} = \frac{11 \times 3}{2 \times 10} = \frac{33}{20}$.

The sum is $\mathbf{\frac{33}{20}}$.

Question 2: Find the sums given below:

(i) $7 + 10 \frac{1}{2} + 14 + \dots + 84$

Solution-

Given $a = 7$, $l = a_n = 84$.
$d = 10\frac{1}{2} - 7 = 3\frac{1}{2} = \frac{7}{2}$.
First, find $n$: $a_n = a + (n – 1) d$.
$84 = 7 + (n - 1) \frac{7}{2}$
$77 = (n - 1) \frac{7}{2}$
$n - 1 = 77 \times \frac{2}{7} = 22$
$n = 23$.
Now find the sum using $S = \frac{n}{2} (a + l)$.
$S_{23} = \frac{23}{2} (7 + 84) = \frac{23}{2} \times 91$
$S_{23} = \frac{2093}{2} = 1046.5$.

The sum is $\mathbf{1046 \frac{1}{2}}$ or 1046.5.

(ii) $34 + 32 + 30 + \dots + 10$

Solution-

Given $a = 34$, $l = a_n = 10$.
$d = 32 - 34 = -2$.
First, find $n$: $a_n = a + (n – 1) d$.
$10 = 34 + (n - 1) (-2)$
$-24 = -2(n - 1)$
$12 = n - 1$
$n = 13$.
Now find the sum $S = \frac{n}{2} (a + l)$.
$S_{13} = \frac{13}{2} (34 + 10) = \frac{13}{2} \times 44$
$S_{13} = 13 \times 22 = 286$.

The sum is 286.

(iii) –5 + (–8) + (–11) + . . . + (–230)

Solution-

Given $a = -5$, $l = a_n = -230$.
$d = -8 - (-5) = -3$.
First, find $n$: $a_n = a + (n – 1) d$.
$-230 = -5 + (n - 1) (-3)$
$-225 = -3(n - 1)$
$n - 1 = \frac{225}{3} = 75$
$n = 76$.
Now find the sum $S = \frac{n}{2} (a + l)$.
$S_{76} = \frac{76}{2} (-5 + (-230)) = 38 \times (-235)$
$S_{76} = -8930$.

The sum is –8930.

Question 3: In an AP:

(i) given $a = 5, d = 3, a_n = 50$, find $n$ and $S_n$.

Solution-

Find $n$ using $a_n = a + (n – 1) d$:
$50 = 5 + (n - 1) 3$
$45 = 3(n - 1)$
$n - 1 = 15 \implies n = 16$.

Find $S_n$ using $S_n = \frac{n}{2} (a + a_n)$:
$S_{16} = \frac{16}{2} (5 + 50) = 8 \times 55 = 440$.

$n = \mathbf{16}$ and $S_n = \mathbf{440}$.

(ii) given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.

Solution-

Find $d$ using $a_{13} = a + 12d$:
$35 = 7 + 12d$
$28 = 12d$
$d = \frac{28}{12} = \frac{7}{3}$.

Find $S_{13}$ using $S_{13} = \frac{13}{2} (a + a_{13})$:
$S_{13} = \frac{13}{2} (7 + 35) = \frac{13}{2} \times 42$
$S_{13} = 13 \times 21 = 273$.

$d = \mathbf{\frac{7}{3}}$ and $S_{13} = \mathbf{273}$.

(iii) given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.

Solution-

Find $a$ using $a_{12} = a + 11d$:
$37 = a + 11(3)$
$37 = a + 33$
$a = 4$.

Find $S_{12}$ using $S_{12} = \frac{12}{2} (a + a_{12})$:
$S_{12} = 6 (4 + 37) = 6 \times 41 = 246$.

$a = \mathbf{4}$ and $S_{12} = \mathbf{246}$.

(iv) given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.

Solution-

Use $a_3 = a + 2d$: $a = 15 - 2d$ (1).

Use $S_{10} = \frac{10}{2} [2a + 9d]$:
$125 = 5 [2a + 9d]$
$25 = 2a + 9d$ (2).
Substitute (1) into (2):
$25 = 2(15 - 2d) + 9d$
$25 = 30 - 4d + 9d$
$5d = 25 - 30 = -5$
$d = -1$.

Find $a_{10}$ using $a_{10} = a + 9d$. First find $a$ using (1):
$a = 15 - 2(-1) = 17$.
$a_{10} = 17 + 9(-1) = 17 - 9 = 8$.

$d = \mathbf{-1}$ and $a_{10} = \mathbf{8}$.

(v) given $d = 5, S_9 = 75$, find $a$ and $a_9$.

Solution-

Find $a$ using $S_9 = \frac{9}{2} [2a + (9 - 1) d]$:
$75 = \frac{9}{2} [2a + 8(5)]$
$75 = \frac{9}{2} [2a + 40] = 9(a + 20)$
$75 = 9a + 180$
$9a = 75 - 180 = -105$
$a = -\frac{105}{9} = -\frac{35}{3}$.

Find $a_9$ using $a_9 = a + 8d$:
$a_9 = -\frac{35}{3} + 8(5) = -\frac{35}{3} + 40$
$a_9 = \frac{-35 + 120}{3} = \frac{85}{3}$.

$a = \mathbf{-\frac{35}{3}}$ and $a_9 = \mathbf{\frac{85}{3}}$.

(vi) given $a = 2, d = 8, S_n = 90$, find $n$ and $a_n$.

Solution-

Find $n$ using $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$90 = \frac{n}{2} [2(2) + (n - 1) 8]$
$180 = n [4 + 8n - 8]$
$180 = n [8n - 4]$
$180 = 8n^2 - 4n$
$8n^2 - 4n - 180 = 0$.
Divide by 4: $2n^2 - n - 45 = 0$.
Factorise: $(2n + 9)(n - 5) = 0$.
Since $n$ must be a positive integer, $n = 5$.

Find $a_n = a_5$ using $a_5 = a + 4d$:
$a_5 = 2 + 4(8) = 2 + 32 = 34$.

$n = \mathbf{5}$ and $a_n = \mathbf{34}$.

(vii) given $a = 8, a_n = 62, S_n = 210$, find $n$ and $d$.

Solution-

Find $n$ using $S_n = \frac{n}{2} (a + a_n)$:
$210 = \frac{n}{2} (8 + 62)$
$210 = \frac{n}{2} (70)$
$210 = 35n$
$n = \frac{210}{35} = 6$.

Find $d$ using $a_n = a_6 = 62$:
$a_6 = a + 5d$
$62 = 8 + 5d$
$54 = 5d$
$d = \frac{54}{5}$.

$n = \mathbf{6}$ and $d = \mathbf{\frac{54}{5}}$ or 10.8.

(viii) given $a_n = 4, d = 2, S_n = –14$, find $n$ and $a$.

Solution-

Use $a_n = a + (n - 1) d$: $4 = a + (n - 1) 2 \implies a = 4 - 2n + 2 = 6 - 2n$ (1).

Use $S_n = \frac{n}{2} (a + a_n)$:
$-14 = \frac{n}{2} (a + 4)$
$-28 = n(a + 4)$ (2).
Substitute (1) into (2):
$-28 = n((6 - 2n) + 4)$
$-28 = n(10 - 2n)$
$-28 = 10n - 2n^2$
$2n^2 - 10n - 28 = 0$.
Divide by 2: $n^2 - 5n - 14 = 0$.
Factorise: $(n - 7)(n + 2) = 0$.
Since $n$ must be positive, $n = 7$.

Find $a$ using (1):
$a = 6 - 2(7) = 6 - 14 = -8$.

$n = \mathbf{7}$ and $a = \mathbf{-8}$.

(ix) given $a = 3, n = 8, S = 192$, find $d$.

Solution-

Find $d$ using $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$S_8 = 192$.
$192 = \frac{8}{2} [2(3) + (8 - 1) d]$
$192 = 4 [6 + 7d]$
$48 = 6 + 7d$
$42 = 7d$
$d = 6$.

$d = \mathbf{6}$.

(x) given $l = 28, S = 144$, and there are total 9 terms. Find $a$.

Solution-

Given $l = a_n = 28$, $S = S_9 = 144$, $n = 9$.
Find $a$ using $S_n = \frac{n}{2} (a + l)$:
$144 = \frac{9}{2} (a + 28)$
$288 = 9(a + 28)$
$32 = a + 28$
$a = 32 - 28 = 4$.

$a = \mathbf{4}$.

Question 4: How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution-

Given $a = 9$, $d = 17 - 9 = 8$, $S_n = 636$. Find $n$.
$S_n = \frac{n}{2} [2a + (n – 1) d]$.
$636 = \frac{n}{2} [2(9) + (n - 1) 8]$
$1272 = n [18 + 8n - 8]$
$1272 = n [10 + 8n]$
$1272 = 8n^2 + 10n$.
$8n^2 + 10n - 1272 = 0$.
Divide by 2: $4n^2 + 5n - 636 = 0$.
Using factorization: $4n^2 + 53n - 48n - 636 = 0$.
$4n(n - 12) + 53(n - 12) = 0$.
$(4n + 53)(n - 12) = 0$.
Since $n$ must be a positive integer, $n = 12$.

The number of terms is 12.

Question 5: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution-

Given $a = 5$, $l = 45$, $S_n = 400$.
Find $n$ using $S_n = \frac{n}{2} (a + l)$:
$400 = \frac{n}{2} (5 + 45)$
$400 = \frac{n}{2} (50)$
$400 = 25n$
$n = \frac{400}{25} = 16$.
Find $d$ using $l = a_n = a + (n – 1) d$:
$45 = 5 + (16 - 1) d$
$40 = 15d$
$d = \frac{40}{15} = \frac{8}{3}$.

The number of terms is 16 and the common difference is $\mathbf{\frac{8}{3}}$.

Question 6: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution-

Given $a = 17$, $l = a_n = 350$, $d = 9$.
Find $n$ using $a_n = a + (n – 1) d$:
$350 = 17 + (n - 1) 9$
$333 = 9(n - 1)$
$n - 1 = \frac{333}{9} = 37$
$n = 38$.
Find the sum $S_{38}$ using $S_n = \frac{n}{2} (a + l)$:
$S_{38} = \frac{38}{2} (17 + 350)$
$S_{38} = 19 (367) = 6973$.

There are 38 terms and their sum is 6973.

Question 7: Find the sum of first 22 terms of an AP in which $d = 7$ and 22nd term is 149.

Solution-

Given $d = 7$, $n = 22$, $a_{22} = 149$.
First find $a$ using $a_{22} = a + 21d$:
$149 = a + 21(7)$
$149 = a + 147$
$a = 2$.
Find the sum $S_{22}$ using $S_n = \frac{n}{2} (a + a_n)$:
$S_{22} = \frac{22}{2} (2 + 149)$
$S_{22} = 11 (151) = 1661$.

The sum of the first 22 terms is 1661.

Question 8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution-

Given $a_2 = 14$, $a_3 = 18$. We need $S_{51}$.
Find the common difference $d$:
$d = a_3 - a_2 = 18 - 14 = 4$.
Find the first term $a$:
$a = a_2 - d = 14 - 4 = 10$.
Find the sum $S_{51}$ using $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$S_{51} = \frac{51}{2} [2(10) + (51 - 1) 4]$
$S_{51} = \frac{51}{2} [20 + 50 \times 4]$
$S_{51} = \frac{51}{2} [20 + 200] = \frac{51}{2} \times 220$
$S_{51} = 51 \times 110 = 5610$.

The sum of the first 51 terms is 5610.

Question 9: If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first $n$ terms.

Solution-

Given $S_7 = 49$ and $S_{17} = 289$.
$S_7 = \frac{7}{2} [2a + 6d] = 7(a + 3d) = 49 \implies a + 3d = 7$ (1).
$S_{17} = \frac{17}{2} [2a + 16d] = 17(a + 8d) = 289 \implies a + 8d = 17$ (2).
Subtracting (1) from (2):
$5d = 17 - 7 = 10$
$d = 2$.
Substitute $d=2$ into (1):
$a + 3(2) = 7 \implies a = 1$.
The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$S_n = \frac{n}{2} [2(1) + (n - 1) 2]$
$S_n = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} [2n] = n^2$.

The sum of the first $n$ terms is $\mathbf{n^2}$.

Question 10: Show that $a_1, a_2, \dots, a_n, \dots$ form an AP where $a_n$ is defined as below: Also find the sum of the first 15 terms in each case.

(i) $a_n = 3 + 4n$

Solution-

Find the first few terms:
$a_1 = 3 + 4(1) = 7$.
$a_2 = 3 + 4(2) = 11$.
$a_3 = 3 + 4(3) = 15$.
The list of numbers is $7, 11, 15, \dots$
Common difference $d = a_2 - a_1 = 11 - 7 = 4$.
Also $a_3 - a_2 = 15 - 11 = 4$.
Since $a_{k+1} - a_k$ is the same, it forms an AP. $a=7, d=4$.
Find $S_{15}$ using $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$S_{15} = \frac{15}{2} [2(7) + (15 - 1) 4]$
$S_{15} = \frac{15}{2} [14 + 14 \times 4] = \frac{15}{2} [14 + 56]$
$S_{15} = \frac{15}{2} = 15 \times 35 = 525$.

The sum of the first 15 terms is 525.

(ii) $a_n = 9 - 5n$

Solution-

Find the first few terms:
$a_1 = 9 - 5(1) = 4$.
$a_2 = 9 - 5(2) = -1$.
$a_3 = 9 - 5(3) = -6$.
The list of numbers is $4, -1, -6, \dots$
Common difference $d = a_2 - a_1 = -1 - 4 = -5$.
Also $a_3 - a_2 = -6 - (-1) = -5$.
Since $a_{k+1} - a_k$ is the same, it forms an AP. $a=4, d=-5$.
Find $S_{15}$ using $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$S_{15} = \frac{15}{2} [2(4) + (15 - 1) (-5)]$
$S_{15} = \frac{15}{2} [8 + 14 \times (-5)] = \frac{15}{2}$
$S_{15} = \frac{15}{2} [-62] = 15 \times (-31) = -465$.

The sum of the first 15 terms is –465.

Question 11: If the sum of the first $n$ terms of an AP is $4n – n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$th terms.

Solution-

Given $S_n = 4n – n^2$.
The first term $a_1$ is equal to $S_1$.
$S_1 = 4(1) - (1)^2 = 4 - 1 = 3$.

The sum of the first two terms is $S_2$:
$S_2 = 4(2) - (2)^2 = 8 - 4 = 4$.

The second term $a_2$ is $S_2 - S_1$:
$a_2 = 4 - 3 = 1$.

The third term $a_3$ is $S_3 - S_2$:
$S_3 = 4(3) - (3)^2 = 12 - 9 = 3$.
$a_3 = 3 - 4 = -1$.

The 10th term $a_{10}$ is $S_{10} - S_9$:
$S_{10} = 4(10) - (10)^2 = 40 - 100 = -60$.
$S_9 = 4(9) - (9)^2 = 36 - 81 = -45$.
$a_{10} = -60 - (-45) = -60 + 45 = -15$.

The $n$th term $a_n$ is $S_n - S_{n-1}$:
$a_n = (4n - n^2) - [4(n - 1) - (n - 1)^2]$
$a_n = (4n - n^2) - [4n - 4 - (n^2 - 2n + 1)]$
$a_n = 4n - n^2 - 4n + 4 + n^2 - 2n + 1$
$a_n = 5 - 2n$.

First term ($S_1$) is 3. Sum of first two terms ($S_2$) is 4. Second term ($a_2$) is 1. 3rd term ($a_3$) is –1. 10th term ($a_{10}$) is –15. $n$th term ($a_n$) is $5 - 2n$.

Question 12: Find the sum of the first 40 positive integers divisible by 6.

Solution-

The positive integers divisible by 6 form an AP: $6, 12, 18, \dots$.
Given $a = 6$, $d = 6$, $n = 40$.
The sum formula is $S_n = \frac{n}{2} [2a + (n – 1) d]$.
$S_{40} = \frac{40}{2} [2(6) + (40 - 1) 6]$
$S_{40} = 20 [12 + 39 \times 6]$
$S_{40} = 20 [12 + 234] = 20 \times 246 = 4920$.

The sum of the first 40 positive integers divisible by 6 is 4920.

Question 13: Find the sum of the first 15 multiples of 8.

Solution-

The multiples of 8 form an AP: $8, 16, 24, \dots$.
Given $a = 8$, $d = 8$, $n = 15$.
The sum formula is $S_n = \frac{n}{2} [2a + (n – 1) d]$.
$S_{15} = \frac{15}{2} [2(8) + (15 - 1) 8]$
$S_{15} = \frac{15}{2} [16 + 14 \times 8]$
$S_{15} = \frac{15}{2} [16 + 112] = \frac{15}{2}$
$S_{15} = 15 \times 64 = 960$.

The sum of the first 15 multiples of 8 is 960.

Question 14: Find the sum of the odd numbers between 0 and 50.

Solution-

The odd numbers between 0 and 50 form an AP: $1, 3, 5, \dots, 49$.
Given $a = 1$, $l = 49$, $d = 2$.
Find $n$ using $a_n = a + (n – 1) d$:
$49 = 1 + (n - 1) 2$
$48 = 2(n - 1)$
$n - 1 = 24$
$n = 25$.
Find the sum $S_{25}$ using $S_n = \frac{n}{2} (a + l)$:
$S_{25} = \frac{25}{2} (1 + 49)$
$S_{25} = \frac{25}{2} (50) = 25 \times 25 = 625$.

The sum of the odd numbers between 0 and 50 is 625.

Question 15: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution-

The penalty amounts form an AP: $200, 250, 300, \dots$
Given $a = 200$, $d = 50$, $n = 30$.
We need to find the total penalty, $S_{30}$, using $S_n = \frac{n}{2} [2a + (n – 1) d]$.
$S_{30} = \frac{30}{2} [2(200) + (30 - 1) 50]$
$S_{30} = 15 [400 + 29 \times 50]$
$S_{30} = 15 [400 + 1450]$
$S_{30} = 15 \times 1850 = 27750$.

The contractor has to pay ₹ 27750 as a penalty.

Question 16: A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Solution-

The prize amounts form an AP with $n = 7$.
Let the first prize be $a$. The common difference $d = -20$.
The total sum $S_7 = 700$.
Use $S_n = \frac{n}{2} [2a + (n – 1) d]$:
$700 = \frac{7}{2} [2a + (7 - 1) (-20)]$
$\frac{700 \times 2}{7} = 2a + 6(-20)$
$200 = 2a - 120$
$2a = 320$
$a = 160$.
The prizes are $a, a+d, a+2d, \dots, a+6d$:
160, $160-20=140$, $140-20=120$, $100, 80, 60, 40$.

The values of the seven prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, and ₹ 40.

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Question 17: In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution-

The number of trees planted by each class is:
Class I: $3 \times 1 = 3$.
Class II: $3 \times 2 = 6$.
Class III: $3 \times 3 = 9$.
...
Class XII: $3 \times 12 = 36$.
The total number of trees forms an AP with $a = 3$, $d = 3$, and $n = 12$.
Find $S_{12}$ using $S_n = \frac{n}{2} (a + l)$, where $l = 36$:
$S_{12} = \frac{12}{2} (3 + 36)$
$S_{12} = 6 (39) = 234$.

A total of 234 trees will be planted by the students.

Question 18: A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii $0.5 \text{ cm}, 1.0 \text{ cm}, 1.5 \text{ cm}, 2.0 \text{ cm}, \dots$ What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$)

Solution-

The lengths of the successive semicircles are $l_1, l_2, l_3, \dots, l_{13}$. The length of a semicircle is $\pi r$.
$l_1 = \pi (0.5)$
$l_2 = \pi (1.0)$
$l_3 = \pi (1.5)$
...
The lengths form an AP with first term $a = 0.5\pi$ and common difference $d = 0.5\pi$. $n = 13$.
We need to find $S_{13}$.
$S_{13} = \frac{13}{2} [2a + (13 - 1) d]$
$S_{13} = \frac{13}{2} [2(0.5\pi) + 12 (0.5\pi)]$
$S_{13} = \frac{13}{2} [\pi + 6\pi] = \frac{13}{2} [7\pi]$
Substitute $\pi = \frac{22}{7}$:
$S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}$
$S_{13} = \frac{13}{2} \times 22 = 13 \times 11 = 143$.

The total length of the spiral is 143 cm.

Question 19: 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution-

The number of logs per row forms an AP: $20, 19, 18, \dots$.
Given $a = 20$, $d = -1$. Total sum $S_n = 200$. Find $n$.
$S_n = \frac{n}{2} [2a + (n – 1) d]$.
$200 = \frac{n}{2} [2(20) + (n - 1) (-1)]$
$400 = n [40 - n + 1]$
$400 = 41n - n^2$
$n^2 - 41n + 400 = 0$.
Factorise: $(n - 16)(n - 25) = 0$.
$n = 16$ or $n = 25$.
If $n=25$, the number of logs in the 25th row ($a_{25}$) is:
$a_{25} = a + 24d = 20 + 24(-1) = -4$. Since the number of logs cannot be negative, $n=25$ is not physically possible.
Thus, $n = 16$.
The number of logs in the top row ($a_{16}$) is:
$a_{16} = a + 15d = 20 + 15(-1) = 5$.

The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

Question 20: In a potato race, a bucket is placed at the starting point, which is $5 \text{ m}$ from the first potato, and the other potatoes are placed $3 \text{ m}$ apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution-

The distance covered to pick up each potato involves running to the potato and running back to the bucket (twice the distance). There are $n=10$ potatoes.
Distance to pick up the 1st potato ($a_1$): $2 \times 5 = 10 \text{ m}$.
Distance to pick up the 2nd potato ($a_2$): $2 \times (5 + 3) = 16 \text{ m}$.
Distance to pick up the 3rd potato ($a_3$): $2 \times (5 + 3 + 3) = 22 \text{ m}$.
This forms an AP: $10, 16, 22, \dots$
Given $a = 10$, $d = 16 - 10 = 6$, $n = 10$.
We need to find the total distance $S_{10}$.
$S_{10} = \frac{10}{2} [2(10) + (10 - 1) 6]$
$S_{10} = 5 [20 + 9 \times 6]$
$S_{10} = 5 [20 + 54]$
$S_{10} = 5 \times 74 = 370$.

The total distance the competitor has to run is 370 m.

EXERCISE 5.4 (Optional)

Question 1: Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find $n$ for $a_n < 0$]

Solution-

The given AP is $121, 117, 113, \dots$.
First term $a = 121$.
Common difference $d = 117 - 121 = -4$.
We need to find the smallest positive integer $n$ such that the $n$th term ($a_n$) is negative, i.e., $a_n < 0$.

Using the formula $a_n = a + (n - 1) d$:
$121 + (n - 1) (-4) < 0$
$121 - 4n + 4 < 0$
$125 - 4n < 0$
$125 < 4n$
$n > \frac{125}{4}$
$n > 31.25$.

Since $n$ must be an integer, the first integer greater than 31.25 is $n = 32$.
We check the 32nd term:
$a_{32} = 121 + (32 - 1) (-4)$
$a_{32} = 121 + 31 \times (-4)$
$a_{32} = 121 - 124 = -3$.
Since $a_{32} = -3$, which is negative, the 32nd term is the first negative term.

The first negative term of the AP is the 32nd term.

Question 2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution-

Let the first term be $a$ and the common difference be $d$.
The terms are given by $a_n = a + (n - 1) d$.

Given that the sum of the 3rd and 7th terms is 6:
$a_3 + a_7 = 6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
$a + 4d = 3$ (Equation 1)

Given that the product of the 3rd and 7th terms is 8:
$a_3 \times a_7 = 8$
$(a + 2d)(a + 6d) = 8$ (Equation 2)

From Equation 1, we express $a$: $a = 3 - 4d$.
Substitute this into Equation 2:
$((3 - 4d) + 2d) ((3 - 4d) + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
$3^2 - (2d)^2 = 8$
$9 - 4d^2 = 8$
$4d^2 = 1$
$d^2 = \frac{1}{4}$
$d = \pm \frac{1}{2}$.

Case 1: If $d = \frac{1}{2}$.
$a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1$.
We need to find the sum of the first 16 terms ($S_{16}$), using $S_n = \frac{n}{2} [2a + (n - 1) d]$.
$S_{16} = \frac{16}{2} \left[2(1) + (16 - 1) \left(\frac{1}{2}\right)\right]$
$S_{16} = 8 \left[2 + 15 \times \frac{1}{2}\right]$
$S_{16} = 8 \left[2 + \frac{15}{2}\right] = 8 \left[\frac{4 + 15}{2}\right] = 8 \times \frac{19}{2} = 4 \times 19 = 76$.

Case 2: If $d = -\frac{1}{2}$.
$a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5$.
$S_{16} = \frac{16}{2} \left[2(5) + (16 - 1) \left(-\frac{1}{2}\right)\right]$
$S_{16} = 8 \left[10 + 15 \times \left(-\frac{1}{2}\right)\right]$
$S_{16} = 8 \left[10 - \frac{15}{2}\right] = 8 \left[\frac{20 - 15}{2}\right] = 8 \times \frac{5}{2} = 4 \times 5 = 20$.

The sum of the first sixteen terms of the AP is 76 or 20.

Question 3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2} \text{ m}$ apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]

Solution-

First, determine the number of rungs ($n$).
Distance between top and bottom rungs $= 2\frac{1}{2} \text{ m} = 2.5 \text{ m} = 250 \text{ cm}$.
Distance between consecutive rungs $= 25 \text{ cm}$.
The number of spaces between rungs is $\frac{\text{Total distance}}{\text{Distance between rungs}} = \frac{250}{25} = 10$.
Since there is one rung at the start and one at the end, the total number of rungs $n$ is (number of spaces) $+ 1$:
$n = \frac{250}{25} + 1 = 10 + 1 = 11$.

The lengths of the rungs form an AP.
First rung length (bottom) $a = 45 \text{ cm}$.
Last rung length (top) $l = 25 \text{ cm}$.
Number of terms $n = 11$.
The total length of wood required is the sum of the lengths of all rungs, $S_{11}$.

Using the sum formula $S_ 1$:
$n = \frac{250}{25} + 1 = 10 + 1 = 11$.

The lengths of the rungs form an AP.
First rung length (bottom) $a = 45 \text{ cm}$.
Last rung length (top) $l = 25 \text{ cm}$.
Number of terms $n = 11$.
The total length of wood required is the sum of the lengths of all rungs, $S_{11}$.

Using the sum formula $S_n = \frac{n}{2} (a + l)$:
$S_{11} = \frac{11}{2} (45 + 25)$
$S_{11} = \frac{11}{2} (70)$
$S_{11} = 11 \times 35 = 385$.

The length of the wood required for the rungs is 385 cm (or $3.85 \text{ m}$).

Question 4: The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$. [Hint : $S_{x – 1} = S_{49} – S_x$]

Solution-

The house numbers form an AP: $1, 2, 3, \dots, 49$.
First term $a=1$, common difference $d=1$.
We are looking for a house numbered $x$, where $1 < x \leq 49$.

The sum of the numbers of houses preceding house $x$ is the sum of the first $(x - 1)$ terms, $S_{x-1}$.
The sum of the numbers of all houses is $S_{49}$.
The sum of the numbers of houses following house $x$ is $S_{49} - S_x$.

The condition given is: Sum of preceding houses = Sum of following houses.
$S_{x-1} = S_{49} - S_x$
$S_{x-1} + S_x = S_{49}$

Using the sum formula $S_n = \frac{n(n+1)}{2}$, since $a=1$ and $d=1$:
$S_{x-1} = \frac{(x - 1) x}{2}$
$S_x = \frac{x(x + 1)}{2}$
$S_{49} = \frac{49(49 + 1)}{2} = \frac{49 \times 50}{2} = 49 \times 25 = 1225$.

Substitute $S_{x-1}$ and $S_x$ into $S_{x-1} + S_x = S_{49}$:
$\frac{(x - 1) x}{2} + \frac{x(x + 1)}{2} = 1225$
$\frac{x}{2} [ (x - 1) + (x + 1) ] = 1225$
$\frac{x}{2} [ 2x ] = 1225$
$x^2 = 1225$
$x = \sqrt{1225} = 35$.

Since $x=35$ is a positive integer between 1 and 49, it is a valid house number.

The value of $x$ is 35.

Question 5: A small terrace at a football ground comprises of 15 steps each of which is $50 \text{ m}$ long and built of solid concrete. Each step has a rise of $\frac{1}{4} \text{ m}$ and a tread of $\frac{1}{2} \text{ m}$. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step $= \frac{1}{4} \times \frac{1}{2} \times 50 \text{ m}^3$]

Solution-

The total number of steps $n = 15$.
The length of each step $L = 50 \text{ m}$.
Rise (height $H$) $= \frac{1}{4} \text{ m}$.
Tread (width $W$) $= \frac{1}{2} \text{ m}$.

The volume of concrete for a step is $\text{Volume} = L \times W \times H$.

The first step: Only one block is formed.
$V_1 = L \times W \times H = 50 \times \frac{1}{2} \times \frac{1}{4}$

The second step: The dimensions (rise and tread) are built upon the first step. The second step adds another layer of the same dimensions, making it $2$ units high (total rise $= 2 \times H$).
$V_2 = L \times W \times (2H) = 50 \times \frac{1}{2} \times 2 \times \frac{1}{4} = 2 V_1$.

The third step: Adds a third unit, making it $3$ units high (total rise $= 3 \times H$).
$V_3 = L \times W \times (3H) = 50 \times \frac{1}{2} \times 3 \times \frac{1}{4} = 3 V_1$.

The volume of concrete in the $n$th step is $V_n = n \times (50 \times \frac{1}{2} \times \frac{1}{4}) \text{ m}^3$.

$V_1 = \frac{50}{8} = \frac{25}{4} \text{ m}^3$.
The sequence of volumes is an AP: $V_1, 2V_1, 3V_1, \dots, 15V_1$.
First term $a = V_1 = \frac{25}{4}$.
Common difference $d = V_1 = \frac{25}{4}$.
Number of terms $n = 15$.

The total volume required is $S_{15}$.
We use $S_n = \frac{n}{2} [2a + (n - 1) d]$:
$S_{15} = \frac{15}{2} \left[2\left(\frac{25}{4}\right) + (15 - 1) \left(\frac{25}{4}\right)\right]$
$S_{15} = \frac{15}{2} \left[\frac{50}{4} + 14 \times \frac{25}{4}\right]$
$S_{15} = \frac{15}{2} \left[\frac{50}{4} + \frac{350}{4}\right]$
$S_{15} = \frac{15}{2} \times \frac{400}{4} = \frac{15}{2} \times 100$
$S_{15} = 15 \times 50 = 750$.

The total volume of concrete required is $750 \text{ m}^3$.