Find the sum of first 22 terms of an AP in which $d = 7$ and 22nd term is 149.

Given:

In an A.P., $d = 7$ and 22nd term is 149.

To do:

We have to find the sum of first 22 terms.

Solution:

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$n$th term $a_n=a+(n-1)d$

This implies,

$a_{22}=a+(22-1)7$

$149=a+21(7)$

$149=a+147$

$a=149-147$

$a=2$

$S_{22}=\frac{22}{2}[2 \times (2)+(22-1) \times 7]$

$=11[4+21 \times 7]$

$=11(4+147)$

$=11 \times (151)$

$=1661$

Therefore, $S_{22}=1661$.