# RS Aggarwal Class 10 Solutions Chapter 11 - Arithmetic Progressions Ex 11E (11.5)

## RS Aggarwal Class 10 Chapter 11 - Arithmetic Progressions Ex 11E (11.5) Solutions Free PDF

Q.1: The sum of the first n terms of an AP is $\left ( \frac{3n^{2}}{2} + \frac{5n}{2} \right )$. Find the nth term and the 25th term of the AP.

Sol:

Let $S_{n}$ denotes the sum of first n terms of the AP.

$S_{n} = \left ( \frac{3n^{2}}{2} + \frac{5n}{2} \right )$

Replacing n by (n -1) in (i), we get:

$S_{n-1} = \frac{1}{2} \left [ 3(n – 1)^{2} + 5(n – 1) \right ]$

$= \frac{1}{2} \left [ 3n^{2} – 6n + 3 + 5n – 5 \right ]$

$= \frac{1}{2} \left [ 3n^{2} – n – 2 \right ]$

nth term of the AP, $a_{n} = S_{n} – S_{n-1}$

$= \frac{1}{2} (3n^{2} + 5n) – \frac{1}{2} \left [ 3n^{2} – n – 2 \right ]$

$= \frac{1}{2} (6n + 2) = 3n + 1$

Putting n = 25 in (ii) we get:

$a_{25} = (3 \times 25) + 1 = 76$

Hence, the nth term is (3n + 1) and 25th term is 76.

Q.2: How many terms of the AP 21, 18, 15, ……… must be added to get the sum 0?

Sol:

The given AP is 21, 18, 15, …….

Here a = 21 and d = 18 – 21 = – 3

Let the required number of terms be n.

Then,

$S_{n} = 0$

$\Rightarrow \frac{n}{2} [2 \times 21 + (n – 1) \times (-3)] = 0$

$\Rightarrow \frac{n}{2} [42 -3n + 3] = 0$

$\Rightarrow n (45 – 3n) = 0$

$\Rightarrow n = 0 \;\; or \;\; 45 – 3n = 0$

$\Rightarrow n = 0 \;\; or \;\; n = 15$

$Therefore, n = 0$       ( Number of terms cannot be 0)

Hence, the required number of terms is 15.

Q.3: How many terms of the AP 9, 17, 25,  ……… must be taken so that their sum is 636?

The given AP is 9, 17, 25,  ………

Here a = 9 and d = 17 – 9 = 8

Let the required number of terms be n.

Then,

$S_{n} = 636$

$\Rightarrow \frac{n}{2} [2 \times 9 + (n – 1) \times 8 ] = 636$

$\Rightarrow \frac{n}{2} [18 + 8n – 8] = 636$

$\Rightarrow \frac{n}{2} (10 + 8n) = 636$

$\Rightarrow n (5 + 4n) = 636$

$\Rightarrow 4n^{2} + 5n – 636 = 0$

$\Rightarrow 4n^{2} – 48 n + 53 n – 636 = 0$

$\Rightarrow 4n(n – 12) + 53 ( n – 12)= 0$

$\Rightarrow (4n + 53 )(n – 12) = 0$

$\Rightarrow 4n + 53 = 0 \;\; or \;\;n – 12 = 0$

$\Rightarrow n = -\frac{53}{4} \;\; or \;\; n = 12$

$Therefore, n = 12$          ( Number of terms cannot be negative).

Hence, the required number of terms is 12.

Q.4: How many terms of the AP 63, 60, 57, 54, ………  must be taken so that their sum is 693? Explain the double answer.

Sol:

The given AP is 63, 60, 57, 54, ………

Here a = 63  and d = 60 – 63 = – 3

Let the required number of terms be n.

Then,

$S_{n} = 693$

$\Rightarrow \frac{n}{2} [2 \times 63 + (n – 1) \times (-3) ] = 693$

$\Rightarrow \frac{n}{2} [126 – 3n + 3] = 693$

$\Rightarrow n (129 – 3n) = 1386$

$\Rightarrow 3n^{2} – 129n + 1386 = 0$

$\Rightarrow 3n^{2} – 66n – 63n + 1386 = 0$

$\Rightarrow 3n(n – 22) – 63 ( n – 22)= 0$

$\Rightarrow (n – 22 )(3n – 63) = 0$

$\Rightarrow n – 22 = 0 \;\; or \;\; 3n – 63 = 0$

$\Rightarrow n = 22 \;\; or \;\; n = 21$

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.

$a_{22} = 63 + (22 – 1) \times (-3) = 63 – 63 = 0$

Hence, the required number of terms is 21 or 22.

Q.5: How many terms of the AP $20, 19\frac{1}{3}, 18\frac{2}{3} , ……….$ must be taken so that sum is 300? Explain the double answer.

Sol:

The given AP is $20, 19\frac{1}{3}, 18\frac{2}{3} , ……….$

Here, a = 20 and d = $19\frac{1}{3} – 20 = – \frac{2}{3}$

Let the required number of terms be n. Then,

$S_{n} = 300$

$\Rightarrow S_{n} = \frac{n}{2} \left [ 2a + (n-1) d \right ]$

$\Rightarrow 300 = \frac{n}{2} \left [ 2 \times 20 + (n-1) \times \left ( – \frac{2}{3} \right ) \right ]$

$\Rightarrow 600 = n \left [ \frac{120 – (2n – 2) }{3} \right ]$

$\Rightarrow 1800 = n (120 – 2n + 2)$

$\Rightarrow 1800 = – 2n^{2} + 122 n$

$\Rightarrow 2n^{2} – 122 n + 1800 = 0$

$\Rightarrow 2n^{2} – 50 n – 72n + 1800 = 0$

$\Rightarrow 2n (n – 25) – 72 (n – 25) = 0$

$\Rightarrow (n – 25) = 0 \;\; or \;\; (2n- 72) = 0$

$\Rightarrow n = 25 \;\; or \;\; n = 36$

So the sum of the first 25 terms as well as that of first 36 terms is 300. This is because the sum of all terms from 26th to 36th is 0.

Hence, the required number of terms is 25 or 36.

Q.6: Find the sum of all odd numbers between 0 and 50.

Sol:

All odd numbers between 0 and 50 are 1, 3, 5, ……, 49.

This is an AP in which a= 1 , d= (3 – 1) = 2 and l = 49

Let the number of terms be n.

Then $T_{n} = 49$

$\Rightarrow a+ (n – 1) d = 49$

$\Rightarrow 1+ (n – 1) 2 = 49$

$\Rightarrow (n – 1) 2 = 48$

$\Rightarrow n = 25$

Therefore, Required sum = $\frac{n}{2} ( a + l)$

$= \frac{25}{2} ( 1 + 49) = \frac{25}{2} (50 )$

$= 25 \times 25 = 625$

Hence, the required sum is 625.

Q.7: Find the sum of all the natural numbers between 200 and 400 which are divisible by 7.

Sol:

The natural number between 200 and 400 which are divisible by 7 are 203, 210, …….. 399.

This is an AP in which a = 203 , d= 7 and l = 399

Let the number of terms be n.

Then $T_{n} = 399$

$\Rightarrow a + (n – 1) d = 399$

$\Rightarrow 203 + (n – 1) 7 = 399$

$\Rightarrow (n – 1) 7 = 196$

$\Rightarrow n = 29$

Therefore, Required sum = $\frac{n}{2} ( a + l)$

$= \frac{29}{2} ( 203 + 399) = \frac{29}{2} (602)$

$= 8729$

Hence, the required sum is 8729.

Q.8: Find the sum of first forty positiveintegersr divisible by 6.

Sol:

The positive integer divisible by 6 are 6, 12, 18, ……

This is an AP with a = 6 and d = 6.

Also, n = 40    (Given)

Using the formula,

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

$\Rightarrow S_{40} = \frac{40}{2} \left [ 2 \times 6 + (40 – 1) \times 6 \right ]$

= 20 ( 12 + 234)

= 20 (246) = 4920

Hence the required sum is 4920.

Q.9: Find the sum of first 15 multiples of 8.

Sol:

First 15 multiples of 8 are 8, 16, 24, …….

This is an AP in which a = 8, d = 16 – 8= 8, n =15

Thus we have

$l = a + (n – 1) d$

$= 8 + (15 – 1) 8$

$l = 120$

Therefore, Required sum = $\frac{n}{2} ( a + l)$

$= \frac{15}{2} (8 + 120) = 15 (64)$

$= 960$

Hence, the required sum is 960.

Q.10: Find the sum of all multiples of 9 lying between 300 and 700.

Sol:

All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, 693

This is an AP in which a = 306, d = (315 – 306) = 9, I = 693

Let the number of these terms be n,

Then $T_{n} = 693$

$\Rightarrow$ a + (n – 1)d=693

$\Rightarrow$ 306 + (n-1) 9 = 693

$\Rightarrow$9 (n – 1) =387

$\Rightarrow$ (n – 1)=43

$\Rightarrow$ n= 44

Required sum=$\frac{n}{2}$ (a+l)

=$\frac{44}{2}$ (306+693)

$\Rightarrow$22 (999)

$\Rightarrow$22 $\times$ (1000-1)

$\Rightarrow$22 $\times$1000 – 22

$\Rightarrow$22000-22 = 21978

Hence, $S_{n}$ = 21978

Q.11: Find the sum of all three-digit natural numbers which are divisible by 13.

Sol:

All three digit natural numbers divisible by 13 are 104, 117, 130, 143,…, 988

This is an AP in which a = 104, d = (117 – 104) = 13, l = 988

Then $T_{n} = 988$

$\Rightarrow$ a + (n – 1)d = 988

$\Rightarrow$ 104 + (n – 1) $\times$ 13= 988

$\Rightarrow$13 $\times$ (n – 1) = 884

$\Rightarrow$ (n – 1) = 68

$\Rightarrow$ n = 69

Required sum = $\frac{n}{2}$(a + l)

= $\frac{69}{2}$ (104 + 988)

$\Rightarrow$69 $\times$ 546

Hence, $S_{n}$ = 37674

Q.12: Find the sum of first 100 even natural numbers which are divisible by 5.

Sol:

First 100 even natural numbers divisible by 5 are 10, 20, 30, … to 100 term

First term of AP, a = 10

Common difference, d = 20 – 10 = 10

Number of terms = n = 100

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

=$\frac{100}{2}$ [2 $\times$ 10 + (100 – 1) $\times$ 10]

= 50(20 + 99 $\times$ 10)

=50( 20 + 990)

= 50 $\times$ 1010 = 50500

Hence, the sum of the first 100 natural numbers which are divisible by 5 is 50500.

Q.13: Find the sum of the following.

$(1-\frac{1}{n}) + (1-\frac{2}{n}) + (1-\frac{3}{n}) + ……..$up to n terms.

Sol:

On simplifying the given series, we get:

$(1-\frac{1}{n}) + (1-\frac{2}{n}) + (1-\frac{3}{n}) + ……..$up to n terms

= $(1 + 1 + 1 + 1 + ……. up\;\; to \;\; n \;\; terms) – \left ( \frac{1}{n} + \frac{2}{n} + \frac{3}{n} +…… + \frac{n}{n}\right )$

$= n – \left ( \frac{1}{n} + \frac{2}{n} + \frac{3}{n} +…… + \frac{n}{n}\right )$

Here $\left ( \frac{1}{n} + \frac{2}{n} + \frac{3}{n} +…… + \frac{n}{n}\right )$ is an AP whose first term is $\frac{1}{n}$ and the common difference is $\frac{2}{n} – \frac{1}{n} = \frac{1}{n}$

$S_{n} = \left [ \frac{n}{2}\left ( 2 \times \frac{1}{n} + \left ( n – 1 \right ) \times \left ( \frac{1}{n} \right )\right ) \right ]$

The sum of n terms of an AP is given by

$= n – \left [ \frac{n}{2}\left ( 2 \times \frac{1}{n} + \left ( n – 1 \right ) \times \left ( \frac{1}{n} \right )\right ) \right ]$

$= n – \left [ \frac{n}{2}\left ( \frac{2}{n} + \left ( \frac{ n – 1}{n} \right )\right ) \right ]$

$= n – \left ( \frac{n + 1}{2} \right ) = \frac{n – 1 }{2}$

Q.14: In an AP it is given that $S_{5} + S_{7} = 167$ and $S_{10}= 235$, then find the AP, where $S_{n}$ denotes the sum of its first n terms.

Sol:

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

$S_{5} + S_{7} = 167$

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

$\Rightarrow \frac{5}{2} (2a + 4d) + \frac{7}{2} (2a + 6d) = 167$

$\Rightarrow 5a + 10 d + 7a + 21 d = 167$

$\Rightarrow 12a + 31 d = 167$         ………….(i)

Also,

$S_{10}= 235$

$\Rightarrow \frac{10}{2} (2a + 9d) = 235$

$\Rightarrow 5 (2a + 9d) = 235$

$\Rightarrow 2a + 9d = 47$

Multiplying both sides by 6, we get

$12a + 54 d = 282$       …………….(ii)

Subtracting (i) from (ii), we get

$12a + 54 d – 12a – 31 d = 282 – 167$

$\rightarrow 23 d = 115$

$\rightarrow d = 5$

Putting d = 5 in (i), we get

$12 a + 31 \times 5 = 167$

$\Rightarrow 12 a + 155 = 167$

$\Rightarrow 12 a = 167 – 155$

$\Rightarrow 12 a = 12$

$\Rightarrow a = 1$

Hence, the AP is 1, 6, 11, 16, …………

Q.15:  In the given AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference.

Sol:

Here, a = 2, l = 29 and $S_{n} = 155$

Let d be the common difference of the given AP and n be the total number of terms.

Then $T_{n} = 29$

$\Rightarrow a + (n – 1)d = 29$

$\Rightarrow 2 + (n – 1)d = 29$          ………….(i)

The sum of n terms of an AP is given by

$S_{n} = \frac{n}{2} (a + l) = 155$

$\Rightarrow \frac{n}{2} (2 + 29) = 155$

$\Rightarrow \frac{n}{2} (31) = 155$

$\Rightarrow n = 10$

Putting the value of n in (i), we get :

$2 + 9d = 29$

$\Rightarrow 9d = 27$

$\Rightarrow d = 3$

Thus the common difference of the given AP is 3.

Q.16:  In the given AP, the first term is -4, the last term is 29 and the sum of all the terms is 150. Find the common difference.

Sol:

Here, a = 2, l = 29 and $S_{n} = 150$

Let d be the common difference of the given AP and n be the total number of terms.

Then $T_{n} = 29$

$\Rightarrow a + (n – 1)d = 29$

$\Rightarrow -4 + (n – 1)d = 29$          ………….(i)

The sum of n terms of an AP is given by

$S_{n} = \frac{n}{2} (a + l) = 150$

$\Rightarrow \frac{n}{2} (-4 + 29) = 150$

$\Rightarrow \frac{n}{2} (25) = 150$

$\Rightarrow n = 12$

Putting the value of n in (i), we get :

$-4 + 11d = 29$

$\Rightarrow 11d = 33$

$\Rightarrow d = 3$

Thus the common difference of the given AP is 3.

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

Sol:

Suppose there are n terms in an AP

Here, a = 17, d = 9 and l = 350

$a_{n} = 350$

$\Rightarrow 17 + (n – 1) \times 9 = 350$

$\Rightarrow 9n + 8 = 350$

$\Rightarrow 9n = 350 – 8 = 342$

$\Rightarrow n = 38$

Thus, there are 38 terms in the AP.

$Therefore, S_{n} = \frac{38}{2} (17 + 350)$

$= 19 \times 367$

= 6973

Hence, the required sum is 6973.

#### Practise This Question

Which disease is caused by exposure to ultra violet radiation?