If the sum of n terms of an A. P. is 3n² + 5n then which of its terms is 164?

Given:

Sum of n terms of an A.P. = 3n² + 5n

Value of one term = 164

Formula Used:

Sum of n terms of an A.P. = n/2 × [2a + (n-1)d]

nth term of an A.P. = a + (n-1)d

Calculation:

The sum of n terms is given as 3n² + 5n.

We know: nth term of an A.P. = sum of n terms - sum of (n-1) terms.

Let the nth term be 164. We need to find n:

Sum of n terms = 3n² + 5n

Sum of (n-1) terms = 3(n-1)² + 5(n-1)

nth term = [3n² + 5n] - [3(n-1)² + 5(n-1)]

164 = [3n² + 5n] - [3(n² - 2n + 1) + 5n - 5]

⇒ 164 = 3n² + 5n - [3n² - 6n + 3 + 5n - 5]

⇒ 164 = 3n² + 5n - 3n² + 6n - 3 - 5n + 5

⇒ 164 = 6n + 2

⇒ 6n = 164 - 2

⇒ 6n = 162

⇒ n = 162 / 6

⇒ n = 27

The 27th term of the A.P. is 164.