# 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 aₙ = 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

a₅₄ = a + (54 - 1)d

= 3 + 53 × 12

= 3 + 636

= 639

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

We get, 132 + 639 = 771

aₙ_{ }= 771

aₙ = 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 54^{th} term.

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

**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

**Summary:**

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

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