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

i) 2, _, 26

ii) _, 13, _ , 3

iii) 5, _ , _ , 9 ½

iv) - 4, _ , _ , _ , _ , 6

v) _ , 38, _ , _ , _ , - 22

**Solution:**

We are given with the Arithmetic Progression and we have to find the missing terms.

(i) 2, _, 26

We know that the n^{th} term of an AP is given by, aₙ = a + (n - 1)d

Let the common difference be d

First term a = 2

Third term = a + (3 - 1)d = a + 2d = 2 + 2d

Third term = 26 (Given)

2 + 2d = 26

2d = 26 - 2

2d = 24

d = 12

Second term = a + d

= 2 + 12

= 14

The missing term in the box is 14.

ii) _, 13, _ , 3

aₙ = a + (n - 1)d

a₂ = 13, a₄ = 3

Second term = 13

a + (2 - 1) d = 13

a + d = 13 ....(1)

Fourth term = 3

a + (4 - 1) d = 3

a + 3d = 3 ....(2)

Solve equations (1) and (2) to find a and d.

On subtracting equation (1) from equation (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

Third term = a + (3 - 1)d = a + 2d

= 18 + 2(-5) = 8

The missing terms in the boxes are 18 and 8 respectively.

iii) 5, _ , _ , 9 ½

aₙ = a + (n - 1)d

First term a = 5

Fourth term, a₄ = 9 ½ = 19/2

Fourth term a₄ = a + (4 - 1) d

5 + 3d = 19/2

6d = 9

d = 3/2

Second term = a₂ = a + d

= 5 + 3/2

= 13/2

Third term, a₃ = a + 2d

= 5 + 2 × (3/2)

= 5 + 3

= 8

The missing terms in the boxes are 13/2 and 8 respectively.

iv) - 4, _ , _ , _ , _ , 6

aₙ = a + (n - 1)d

First term a = - 4

Sixth term a₆ = 6

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

- 4 + 5d = 6

5d = 10

d = 2

Second term a + d = - 4 + 2 = - 2

Third term a + 2d = - 4 + 4 = 0

Fourth term a + 3d = - 4 + 6 = 2

Fifth term a + 4d = - 4 + 8 = 4

The missing terms are - 2, 0, 2, 4.

v) _ , 38, _ , _ , _ , - 22

aₙ = a + (n - 1)d

Let the first term be = a

Common difference = d

Second term = 38 (Given)

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

Sixth term = - 22 (Given)

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

On Subtracting equation (1) from equation (2) for a and d, we get

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

4d = - 60

d = - 15

Put the value of d in equation (1)

a - 15 = 38

a = 53

Third term = a + (3 - 1)d = a + 2d

= 53 + 2(-15) = 23

Fourth term = a + 3d

= 53 + 3(-15) = 8

Fifth term = a + 4d

= 53 + 4(-15) = - 7

The missing terms are 53, 23, 8, - 7.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## In the following AP, find the missing terms in the boxes: i) 2, _, 26 ii) _, 13, _ , 3 iii) 5, _ , _ , 9 ½ iv) - 4, _ , _ , _ , _ , 6 v) _ , 38, _ , _ , _ , - 22

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2 Question 3

**Summary:**

In the following AP, the missing terms in the boxes: i) 2, _, 26 ii) _, 13, _ , 3 iii) 5, _ , _ , 9 ½ iv) - 4, _ , _ , _ , _ , 6 v) _ , 38, _ , _ , _ , - 22 are i) 14 ii) 18 and 8 iii) 13/2 and 8 iv) -2, 0, 2 and 4 v) 53, 23, 8 and -7

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