**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 \(n^{th}\)

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 10*^{th}* 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:**

\(\frac{4}{5}\)

Therefore, a – \(\frac{4}{5}\)

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

\(\Rightarrow\)

Therefore, p = \(\frac{28}{7}\)

**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

\(\Rightarrow\)

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 \(S_{p}\)

\(Therefore, S_{p}=ap^{2}+bp\)

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

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

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

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*** \(\left ( 3n^{2}+5n \right )\)**

**Solution:**

Let \(S_{n}\)

\(Therefore, S_{n}=3n^{2}+5n\)

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

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

\(=3n^{2}-n-2\)

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 4*^{th}* term is 9 and the sum of its 6*^{th}* and 13*^{th}* 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,

\(a_{6}+a_{13}=40\)

=> (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

S_{n} = \(\frac{n}{2}\)

\(S_{19}\)

Since, a = 2, d = 5, n =19]

= \(\frac{19}{2}\)

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

S_{n} = \(\frac{n}{2}\)

\(S_{14}\)

= \( 7 (18 – 26 ) \)

= – 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

S_{n} = \(\frac{n}{2}\)

\(S_{12}\)

= 6 \(\times\)

= 6 \(\times\)

= – 180

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

**(iv)** Here a = \(\frac{1}{15} \)

Using the formula

S_{n} = \(\frac{n}{2}\)

\(S_{11}\)

= \(\frac{11}{2} \times \frac{18} {60} \)

= \( \frac{33}{20} \)

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

S_{n} = \(\frac{n}{2}\)

\(S_{100}\)

= 50 \(\times\)

= 50 \(\times\)

= 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 = \( 10\frac{1}{2} – 7 = \frac{7}{2} \)

Let the given series contain n terms. Then,

\(a_{n}\)

\(\Rightarrow 7 + (n – 1) \times \frac{7}{2} = 84\)

\(\Rightarrow \frac{7}{2} n + \frac{7}{2} = 84\)

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

\(\Rightarrow n = \frac{161}{7} = 23\)

Therefore, Required sum = \(\frac{23}{2} \times (7 + 84)\)

\(= \frac{23}{2} \times 91\)

\(= \frac{2093}{2}\)

\(= 1046 \frac{1}{2}\)

**(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,

\(a_{n}\)

\(\Rightarrow 34 + (n – 1) \times (-2) = 10 \)

\(\Rightarrow \; -2n = 10 – 36 = -26 \)

\(\Rightarrow n = 13 \)

\(\Rightarrow n = \frac{161}{7} = 23\)

Therefore, Required sum = \(\frac{13}{2} \times (34 + 10 ) \)

\( = \frac{13}{2} \times 44 \)

\( = 286 \)

**(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,

\(a_{n}\)

\(\Rightarrow -5 + (n – 1) \times (-3) = -230 \)

\(\Rightarrow \; -3n – 2 = -230 \)

\(\Rightarrow -3n = -228 \)

\(\Rightarrow n = 76 \)

Therefore, Required sum = \(\frac{76}{2} \times [ (-5) + (- 230 ) \)

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

\( = -8930 \)

**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 \(a_{n}\)

\(Therefore, a_{n}\)

Putting n = 1, we get

First term, a = \(a_{1}\)

Putting n = 2, we get

\(a_{2} \)

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,

S_{n} = \(\frac{n} {2} \)

= \(\frac{n} {2} \)

\(\frac{n}{2} (-2 – 6n + 6)\)

= 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 \(S_{n} \)

\(Therefore, S_{n} = 3n^{2} + 6n\)

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

\(= 3(n^{2} – 2n + 1)+ 6(n – 1)\)

= \(= 3n^{2} – 3\)

\(Therefore, \)

\(a_{n} = S_{n} – S_{n-1} \)

= \((3n^{2} + 6n) – (3n^{2} – 3)\)

= 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 : \(S_{n} = (3n^{2} – n)\)

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)** 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:**

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

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

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

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

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

\(T_{n} = S_{n} – S_{n-1}\)

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

\(= \frac{1}{2} (10n – 2) = 5n – 1\)

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.