Question

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

Answer 1

Answer:

$S_{30}=2940$.

Step-by-step 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 : $S_{n} =\frac{n}{2}[2a+(n-1)d]$.

d = common difference = 6.

For  n = 30 .

$S_{30} =\frac{30}{2}[2(11+(30-1)6]$.

$S_{30} =15[22+29 *6]$.

$S_{30} =15[22+29 *6]$.

$S_{30} =15[22+174]$.

$S_{30} =15[196]$.

$S_{30}=2940$.

Therefore, $S_{30}=2940$.

Answer 2

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