# Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54^{th} term?

**Solution:**

The formula for n^{th} term of an arithmetic progression is a_{n} = a + (n - 1) d .

Here, an is the nth term, a is the first term, d is the common difference and n is the number of terms.

Given AP is 3, 15, 27, 39.

First term a = 3

Second term a + d = 15

d = 15 - 3 = 12

54^{th} term of the AP is

a_{54} = a + (54 - 1)d

= 3 + 53 × 12

= 3 + 636

= 639

Let n^{th} term of AP be 132 more than 54^{th} term (Given)

We get, 132 + 639 = 771

a_{n }= 771

a_{n} = a + (n - 1)d

771 = 3 + (n - 1)12

768 = (n - 1)12

(n - 1) = 64

n = 65

Therefore, the 65^{th} term will be 132 more than the 54th term.

Alternatively,

Let n^{th} term be 132 more than the 54^{th} term.

n = 54 + 132/12

= 54 + 11

= 65^{th} Term

Answer: The 65^{th} term will be 132 more than the 54th term.

**Video Solution:**

## Which term of the AP 3,15,27,39….. will be 132 more than its 54^{th} term?

### NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2 Question 11 - Chapter 5 Exercise 5.2 Question 11:

Which term of the AP 3,15,27,39….. will be 132 more than its 54^{th} term?

The 65th term of the AP 3,15,27,39 will be 132 more than its 54th term