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In an AP, if Sₙ = 3n² + 5n and ak = 164, find the value of k

Solution:

Given, the expression for the sum of the series is Sₙ = 3n² + 5n

Last term, l is a_k = 164

We have to find the value of k.

Put n = 1, S₁ = 3(1)² + 5(1) = 3 + 5 = 8

S₂ = 3(2)² + 5(2) = 3(4) + 10 = 12 + 10 = 22

The AP in terms of common difference is given by

a, a+d, a+2d, a+3d,......., a+(n-1)d

So, S₁ = a

First term, a = 8

S₂ = sum of first two term of an AP

= a+ a + d

= 2a + d

To find the common difference d,

2a + d = 22

2(8) + d = 22

16 + d = 22

d = 22 - 16

d = 6

The series can be framed as

a = 8

a + d = 8 + 6 = 14

a + 2d = 8 + 2(6) = 8 + 12 = 20

a + 3d = 8 + 3(6) = 8 + 18 = 26

Therefore, the series is 8, 14, 20, 26,.....

The nth term of the series in AP is given by

aₙ = a + (n - 1)d

164 = 8 + (n - 1)(6)

164 - 8 = 6(n - 1)

n - 1 = 156/6

n - 1 = 26

n = 26 + 1

n = 27

Therefore, the value of k is 27.

✦ Try This: In an AP, if Sₙ = n² + 3n, find the nth term of the AP

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 25

In an AP, if Sₙ = 3n² + 5n and a_k = 164, find the value of k

Summary:

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. In an AP, if Sₙ = 3n² + 5n and a_k = 164, then the value of k is 27

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