Question

The sum of n terms of an A.P. is 3n² + 5n, then 164 is its

Options

•  24^th term

•  27^th term

• 26^th term

•  25^th term

Answer 1

Here, the sum of first n terms is given by the expression,

 `S_n = 3n^2 + 5n`

We ned to find which term of the A.P. is 164.

Let us take 164 as the n^th term.

So we know that the n^thterm of an A.P. is given by,

`a_n = S_n - S_( n-1) `

So,

`164 = S_n - S_( n-1) ` 

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

Using the property,

`( a - b)^2 =  a^2 + n^2 - 2ab`

We get,

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

`164 = 3n^2 + 5n - [ 3n^2 + 3 - 6n + 5n - 5 ]`

`164 = 3n^2 + 5n - (3n^2  - n - 2)`

`164 = 3n^2 + 5n  - 3n^2 + n + 2 `

164 = 6n + 2 

Further solving for n, we get

6n = 164 - 2 

`  n = 162/6`

   n = 27

Therefore,  164 is the 27^th term of the given A.P.