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

(i) 

2, __, 26

We know, 

a_n= a+ (n -1) d

Here 

a₃=26, a = 2

26 = 2+ (3-1) d

d = 12

Therefore second term = 12+2 = 14

(ii)

 __,13, __, 3

We know, 

a_n= a+ (n -1) d

Here 

a₄=3, a₂ = 13

3 = a + (4-1) d

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

And 

13= a + 2d --(2) 

Solving 1 and 2 we get 

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

3d - d = 3 - 13

2d = - 10

d = - 5

Putting d = -5 in equation (1)

a + (- 5) = 13

a - 5 = 13

a = 18

3rd term = 13 - 5 = 8

Therefore third term = 8

(iii)

5,__,__,9$\frac{1}{2}$

aₙ = a + (n - 1)d

First term a = 5

 a₄ = 9 ½ = 19/2

 a₄ = a + (4 - 1) d

5 + 3d = 19/2

6d = 9

d = 3/2

 a₂ = a + d

= 5 + 3/2

= 13/2

 a₃ = a + 2d

= 5 + 2 × (3/2)

= 5 + 3

= 8

Therefore, the second and third terms are 13/2 and 8 respectively.

iv) - 4, _ , _ , _ , _ , 6

aₙ = a + (n - 1)d

 a = - 4,  a₆ = 6

a + (6 - 1)d = a + 5d = 6

- 4 + 5d = 6

5d = 10

d = 2

Therefore the required terms are as follows :

 a + d = - 4 + 2 = - 2

 a + 2d = - 4 + 4 = 0

a + 3d = - 4 + 6 = 2

 a + 4d = - 4 + 8 = 4

v) _ , 38, _ , _ , _ , - 22

aₙ = a + (n - 1)d

Let the first term be = a

Common difference = d

a₂ = 38 (Given)

a + (2 - 1)d = a + d = 38 ....(1)

a₆ = - 22 (Given)

a + (6 - 1)d = a + 5d = - 22 ....(2)

Solving equation 1 and 2

a + 5d - (a + d) = -2 2 - 38

4d = - 60

d = - 15

Put the value of d in equation (1)

a - 15 = 38

a = 53

Therefore the required terms are : 

 a + (3 - 1)d = a + 2d

= 53 + 2(-15) = 23

 a + 3d

= 53 + 3(-15) = 8

 a + 4d

= 53 + 4(-15) = - 7