# RS Aggarwal Solutions Class 10 Ex 11D

## RS Aggarwal Class 10 Ex 11D Chapter 11

Question-1: Which term of the AP 21, 18, 15, … is zero?

Solution:

The given AP is 21 ,  18 ,  15 , ….

First term = 21 ,  common difference =  18  –  21 =  – 3

Let $n^{th}$  term be zero

a + (n – 1)d = 0  or  21 + (n – 1)(-3) = 0

21 – 3n + 3 = 0

3n = 24 or n = 8

Hence, 8th term of given series is 0.

Question-2: Find the sum of first n natural numbers.

Solution:

Sum of n natural numbers = 1 + 2 + 3 +   …..  + n

Here a = 1 ,  d = 2 – 1 = 1

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

Therefore, Sum of natural numbers $\frac{n}{2}[ 2 X 1 + (n-1)d]$

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

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

Question-3: Find the sum of first n even natural numbers.

Solution:

Sum of even natural numbers = 2 + 4 + 6 + … to n terms

a = 2 ,  d = 4 – 2 = 2

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

Therefore,  Sum of even numbers

=  $\frac{n}{2}[ 2 X 2 + (n-1) X 2]$

= $\frac{n}{2}[ 4 + 2n- 2 ]$

= $\frac{n}{2}[ (2n + 2)]$

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

= n(n +1) .

Question-4: The first term of an AP is p and its common difference is q. Find its 10th term.

Solution:

First term of AP = a = p

Common difference = d = q

nth term = a + (n – 1)d

10th term = p + (10 – 1)q = p + 9q

Question-5: If 4/5, a, 2 are in AP, find the value of a.

Solution:

$\frac{4}{5}$ ,  a , 2  are in AP .

Therefore,  a  – $\frac{4}{5}$  =  2 –  a  or  2a  =  2 + $\frac{4}{5}$  = $\frac{14}{5}$

a = $\frac{7}{5}$

Question-6: If (2p+1), 13, (5p-3) are in AP, find the value of p.

Solution:

(2p+ 1) ,  13 ,  (5p – 3) are in AP

Therefore,   13-(2p+1) = ( 5p-3)-13  or  26 = 2p+ 1+ 5p – 3

$\Rightarrow$  26 = 7p – 2  or  7p = 28 p

Therefore,   p = $\frac{28}{7}$ = 4.

Question-7: If (2p-1), 7, 3p are in AP, find the value of p.

Solution:

(2p – 1), 7, 3p are in AP

Therefore,   7 – (2p – 1) = 3p – 7 or 7-2p + 1=3p – 7

$\Rightarrow$ 5p = 15

Therefore,  p= 3.

Question-8: If the sum of first p terms of an AP is $\left ( ap^{2}+bp \right )$, find its common difference.

Solution:

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

$Therefore, S_{p}=ap^{2}+bp$

$\Rightarrow S_{p-1}=a\left ( p-1 \right )^{2}+b\left ( p-1 \right )$

$=a\left ( p^{2}-2p+1 \right )+b\left ( p-1 \right )$

$=ap^{2}-\left ( 2a-b \right )p+\left ( a-b \right )$

Now,

$p^{th}\:term\:of\:the\:AP,a_{p}=S_{p}-S_{p-1}$

$=\left ( ap^{2}+bp \right )-\left [ ap^{2}-\left ( 2a-b \right )p+\left ( a-b \right ) \right ]$

$=ap^{2}+bp-ap^{2}+\left ( 2a-b \right )p-\left ( a-b \right )$

= 2ap – (a-b)

Let d be the common difference of the AP.

$Therefore, d=a_{p}-a_{p-1}$

$=\left [ 2ap-\left ( a-b \right ) \right ]-\left [ 2a\left ( p-1 \right )-\left ( a-b \right ) \right ]$

= 2ap – (a-b) – 2a (p-1) + (a-b)

= 2a

Hence, the common difference of the AP is 2a.

Question-9: If the sum of first n terms is $\left ( 3n^{2}+5n \right )$, find its common difference.

Solution:

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

$Therefore, S_{n}=3n^{2}+5n$

$\Rightarrow S_{n-1}=3\left ( n-1 \right )^{2}+5\left ( n-1 \right )$

$=3\left ( n^{2}-2n+1 \right )+5\left ( n-1 \right )$

$=3n^{2}-n-2$

Now,

$n^{th}\:term\:of\:the\:AP,a_{n}=S_{n}-S_{n-1}$

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

= 6n+2

Let d be the common difference of the AP.

$Therefore, d=a_{n}-a_{n-1}$

= (6n+2) – [6(n-1) +2]

= 6n + 2 – 6(n-1) – 2

= 6

Hence, the common difference of the AP is 6.

Question-10: Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Solution:

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

$a_{4}=9$

=> a + (4-1)d = 9                $\left [ a_{n}=a+\left ( n-1 \right )d \right ]$

=> a + 3d = 9   ……(1)

Now,

$a_{6}+a_{13}=40$       (Given)

=> (a+5d) + (a+12d) = 40

=> 2a + 17d = 40    ……(2)

From (1) and (2), we get

2(9-3d) + 17d = 40

=> 18-6d+17d = 40

=> 11d = 40-18 = 22

=> d = 2

Putting d = 2 in (1), we get

a + 3 x 2 = 9

=> a = 9-6 = 3

Hence, the AP is 3, 5, 7, 9, 11,….

## Exercise – 11D

Question 1: Find the sum of each of the following A.P.

(i) 2 , 7 , 12 , 17 ………..to 19 terms

(ii) 9 , 7 , 5 , 3 , ………… to 14 terms

(iii) -37 , -33 , -29 , …….. to 12 terms

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

(v) 0.6 , 1.7 , 2.8 , …….. to 100 terms

(i) Here a = 2, d = (7 – 2) = 5, and n = 19

Using the formula

Sn = $\frac{n}{2}$ [2a+ (n – 1) d], we get

$S_{19}$ = $\frac{19}{2}$[2X2+(19 – 1) x 5]

Since,  a = 2, d = 5, n =19]

= $\frac{19}{2}$ (4 + 90)= 893

Hence, the sum of first 19 terms of the given AP is 893

(ii) Here a = 9, d = (7 – 9) = – 2, and n = 14

Using the formula

Sn = $\frac{n}{2}$ [2a+ (n – 1)d], we get

$S_{14}$ = $\frac{14}{2}$[2 $\times$ 9 + (14 – 1) $\times$ -2]

= $7 (18 – 26 )$

= – 56

Hence, the sum of first 14 terms of the given AP is – 56

(iii) Here a = – 37, d = (-33  – (-37)) = 4 , and n = 12

Using the formula

Sn = $\frac{n}{2}$ [2a+ (n – 1)d], we get

$S_{12}$ = $\frac{12}{2}$[2 $\times$ (- 37) + (12 – 1) $\times$ 4]

= 6 $\times$ (- 77 + 44)

= 6 $\times$ (- 30)

= – 180

Hence, the sum of first 12 terms of the given AP is – 180

(iv) Here a = $\frac{1}{15}$ , d =  $\frac{1}{12} – \frac{1}{15} = \frac{1}{60}$ , and n = 11

Using the formula

Sn = $\frac{n}{2}$ [2a+ (n – 1)d], we get

$S_{11}$ = $\frac{11}{2}$[ 2 $\times$ ($\frac{1}{15}$ ) + (11 – 1) $\times \; \frac{1}{60}$ ]

= $\frac{11}{2} \times \frac{18} {60}$

= $\frac{33}{20}$

Hence, the sum of first 11 terms of the given AP is $\frac{33}{20}$.

(v) Here a = 0.6 , d = 1.7 – 0.6 = 1.1 , and n = 100

Using the formula

Sn = $\frac{n}{2}$ [2a+ (n – 1)d], we get

$S_{100}$ = $\frac{100} {2}$[2 $\times$ (0.6) + (100 – 1) $\times$ 1.1 ]

= 50 $\times$ (1.2 + 108.9)

= 50 $\times$ 110.1

= 5505

Hence, the sum of first 100 terms of the given AP is 5505

Question 2: Find the sum of each of the following arithmetic series

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

(ii) 34 + 32 + 30+ ……..+ 10

(iii) (-5) + (-8) + (-11) + …….(-230)

Sol:

(i) The given arithmetic series is $7 + 10\frac{1}{2} + 14 + …….+ 84$ .

Here a = 7, d = $10\frac{1}{2} – 7 = \frac{7}{2}$ and l = 84

Let the given series contain n terms. Then,

$a_{n}$ = 84

$\Rightarrow 7 + (n – 1) \times \frac{7}{2} = 84$                  [ $a_{n} = a + (n – 1)d$ ]

$\Rightarrow \frac{7}{2} n + \frac{7}{2} = 84$

$\Rightarrow \frac{7}{2} n = 84 – \frac{7}{2} = \frac{161}{2}$

$\Rightarrow n = \frac{161}{7} = 23$

Therefore, Required sum = $\frac{23}{2} \times (7 + 84)$                        [ $S_{n} = \frac{n}{2}(a + l)$ ]

$= \frac{23}{2} \times 91$

$= \frac{2093}{2}$

$= 1046 \frac{1}{2}$

(ii) The given arithmetic series is 34 + 32 + 30+ ……..+ 10

Here a = 34, d = 32 – 34 = -2 and l = 10

Let the given series contain n terms. Then,

$a_{n}$ = 10

$\Rightarrow 34 + (n – 1) \times (-2) = 10$                  [ $a_{n} = a + (n – 1) d$ ]

$\Rightarrow \; -2n = 10 – 36 = -26$

$\Rightarrow n = 13$

$\Rightarrow n = \frac{161}{7} = 23$

Therefore, Required sum = $\frac{13}{2} \times (34 + 10 )$                        [ $S_{n} = \frac{n}{2} (a + l)$ ]

$= \frac{13}{2} \times 44$

$= 286$

(iii) The given arithmetic series is (-5) + (-8) + (-11) + …….(-230)

Here a = -5, d = -8 – (-5) = -3 and l = 230

Let the given series contain n terms. Then,

$a_{n}$ = -230

$\Rightarrow -5 + (n – 1) \times (-3) = -230$                  [ $a_{n} = a + (n – 1) d$ ]

$\Rightarrow \; -3n – 2 = -230$

$\Rightarrow -3n = -228$

$\Rightarrow n = 76$

Therefore, Required sum = $\frac{76}{2} \times [ (-5) + (- 230 )$ ]                        [ $S_{n} = \frac{n}{2} (a + l)$ ]

$= \frac{76}{2} \times (-235)$

$= -8930$

Question 3: Find the sum of n terms of an AP whose nth term is (5 – 6n). Hence find the sum of its first 20 terms.

Sol:

Let $a_{n}$ be the nth term of the AP.

$Therefore, a_{n}$ = 5 – 6n

Putting n = 1, we get

First term, a = $a_{1}$ = 5 – $6 \times 1 = – 1$

Putting n = 2, we get

$a_{2}$ = 5 – $6 \times 2 = – 7$

Let d be the common difference of the AP.

$Therefore, d = a_{2} – a_{1} = -7 – (-1) = -6$

Sum of first n terms of the AP,

Sn = $\frac{n} {2}$ [ 2a + (n – 1)d ]

= $\frac{n} {2}$ [ 2 $\times$ (-1) + (n – 1) $\times$ (-6) ]

$\frac{n}{2} (-2 – 6n + 6)$

= n (2 – 3n)

= 2n – $3n^{2}$

Putting n = 20, we get

$S_{20} = 2 \times 20 – 3 \times 20^{2} = 40 – 1200 = -1160$

Question 4: The sum of the first n terms of an AP is $(3n^{2} + 6n)$. Find the nth term and the 15th term of this AP.

Sol:

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

$Therefore, S_{n} = 3n^{2} + 6n$

$\Rightarrow S_{n-1} = 3(n-1)^{2} + 6(n-1)$

$= 3(n^{2} – 2n + 1)+ 6(n – 1)$

= $= 3n^{2} – 3$

$Therefore,$ nth term of the AP,

$a_{n} = S_{n} – S_{n-1}$

= $(3n^{2} + 6n) – (3n^{2} – 3)$

= 6n + 3

Putting n = 15, we get

$a_{15} = 6 \times 15 + 3 = 90 + 3 = 93$

Hence, the nth term is (6n + 3) and 15th term is 93.

Q.5 : The sum of the first n terms of an AP is given by $S_{n} = (3n^{2} – n)$. Find its

(i) nth term

(ii) First term

(iii) Common difference

Sol:

Given :  $S_{n} = (3n^{2} – n)$              ……….(i)

Replacing n by (n – 1) in (i) we have,

$S_{n-1} = 3(n-1)^{2} – (n-1)$

$= 3(n^{2} -2n + 1) – n + 1$

$= 3n^{2} – 6n + 3 – n + 1$

$= 3n^{2} – 7n + 4$

(i) Now, $T_{n} = S_{n} – S_{n-1}$

$= (3n^{2} – n) – (3n^{2} – 7n + 4) = 6n – 4$

Therefore, nth term, $T_{n} = (6n – 4)$        ………(ii)

(ii) Putting n = 1 in (ii), we get :

$T_{1} = (6 \times 1 )- 4 = 2$

(iii) Putting n = 2 in (ii), we get :

$T_{2} = (6 \times 2 )- 4 = 8$

Therefore, Common difference, $d = T_{2} – T_{1} = 8 – 2 = 6$

Q.6: The sum of the first n terms of an AP is $\left ( \frac{5n^{2}}{2} + \frac{3n}{2} \right )$. Find the nth term and the 20th term of the AP.

Sol:

$S_{n} = \left ( \frac{5n^{2}}{2} + \frac{3n}{2} \right ) = \frac{1}{2} (5n^{2} + 3n)$        ……… (i)

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

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

$= \frac{1}{2} \left [ 5n^{2} – 10n + 5 + 3n – 3 \right ]$

$= \frac{1}{2} \left [ 5n^{2} – 7n + 2 \right ]$

$T_{n} = S_{n} – S_{n-1}$

$= \frac{1}{2} (5n^{2} + 3n) – \frac{1}{2} \left [ 5n^{2} – 7n + 2 \right ]$

$= \frac{1}{2} (10n – 2) = 5n – 1$               ………(ii)

Putting n = 20 in (ii) we get:

$T_{20} = (5 \times 20) – 1 = 99$

Hence, the nth term is (5n – 1) and 20th term is 99.

#### Practise This Question

The element with least ionization energy in group 1 is Caesium.