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

By BYJU'S Exam Prep

Updated on: September 25th, 2023

To find: Term more than 132 of the given AP: 3, 15, 27, 39,….

Formula to find the nth term of an AP: a_n = a + (n – 1) d.

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

In the given AP: 3, 15, 27, 39,….

First term (a) = 3,

Second term (a+d) = 15

Common difference (d) = 15 – 3 = 2

Now, we will find the 54th term of the given AP

a₅₄ = a + (n – 1) d

a₅₄ = 3 + (54 -1) 2

a₅₄ = 3 + 53 × 12

a₅₄ = 3 + 636

a₅₄ = 639

From the question statement, the nth term of AP is 132 more than the 54th term.

So, a_n = 132 + 639 = 771

a_n = a + (n – 1) d

771 = 3 + (n – 1)12

771 = 3 + 12 n – 12

771 – 3 + 12 = 12n

780 = 12n

n = 780/12

n = 65

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

Table of content

1. Term of AP 3, 15, 27, 39 that will be 132 more than its 54th term
2. Which term of the AP 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Term of AP 3, 15, 27, 39 that will be 132 more than its 54th term

An arithmetic progression (AP) in mathematics is a list or sequence of numbers where each term is obtained by adding a fixed number to the term before it. The fixed number is referred to as the arithmetic progression’s common difference.

The fixed number may be positive, negative, and zero. In an AP, the first term is represented by a₁, second term is represented by a₂, and similarly the nth term is represented a_n.

To understand this in a better way, let us take help of an example.

Let n = 10 for an AP 2, 4, 6, 8,….

As discussed, the nth term of an AP: a_n= a + (n – 1) d.

a₁₀ = 2 + (10 – 1)2

a₁₀ = 2 + 20 – 2

a₁₀ = 22 – 2

a₁₀ = 20

So, the 10th term of AP 2, 4, 6, 8,…. is 20.

Summary: