Question

a=2,d=3, find S₃₀,For the given AP, find

Answer 1

In the attachment I have answered this problem.

Formula:

Sum of first 'n ' terms of an AP is

Sn = (n/2)[2a+(n-1)d]

See the attachment for detailed solution

Answer 2

Given, first term , a = 2 and common difference , d = 3.

use sum of n terms in AP ,
$\bf{S_n=\frac{n}{2}[2a+(n-1)d]}$

$\bf{S_{30}=\frac{30}{2}[2a+(30-1)d]}\\\\=15[2a+29d]$

put a = 2 and d = 3

$S_{30}=15[2\times2+29\times3]$

= 15 × [4 + 87]

= 15 × 91

= 1365

hence, $S_{30}=1365$