Question

The sum of the first 30 terms of the sequence an=6n+5 is

Answer 1

First term a1 = 6+5 = 11
second term = 12+5 = 17

common difference = 6

Sum of n terms:-
Sn = (n/2) [ 2a1 + d(n - 1)]
Sum of 30 terms:-
S30 = (30/2)[ 2*11 + 6(30-1)]
= 15 * ( 22 + 174)
= 15 * 196
= 2940 Answer

Answer 2

`S_(30)=2940`.

Explanation:

Given :
`a_(n) =6n+5`.

To find : The sum of the first 30 terms .

Solution: We have given
`a_(n) =6n+5`.

For n = 1

`a_(1) =6(1)+5`.

`a_(1) =6+5`.

`a_(1) =11`.

For n =2

`a_(2) =6(2)+5`.

`a_(2) =17`.

Common difference = 17 - 11 = 6.

Sum of nth term :
`S_(n) =(n)/(2)[2a+(n-1)d]`.

d = common difference = 6.

For n = 30 .

`S_(30) =(30)/(2)[2(11+(30-1)6]`.

`S_(30) =15[22+29 *6]`.

`S_(30) =15[22+29 *6]`.

`S_(30) =15[22+174]`.

`S_(30) =15[196]`.

`S_(30)=2940`.

Therefore,
`S_(30)=2940`.