Question

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

Answer 1

n=22
d=7
a22= 149= l (last term)
a=?

a+(n-1)d= an
a+21*7=149
a+147= 149
a= 2


so, S=n/2(a+l)
= 22/2 (2+149)
=11 *151
=1661.

It is your answer. please check it with the answer provided in the book

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let \:a\: be\: first\: term\: be\: a\: and\: d\: be\: Common\: difference$

$\bf\huge d = 7 \:and\: a_{22} = 149$

$\bf\huge => a + (n - 1)d = 149$

$\bf\huge => a + 21\times 7 = 149$

$\bf\huge => a = 149 - 147 = 2$

$\bf\huge Substitute n = 22 , a = 2\: and\: d = 7$

$\bf\huge S_{n} = \frac{N}{2} [2a + (n - 1)d]$

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

$\bf\huge = 11(4 + 21\times 7)$

$\bf\huge = 11(4 + 147)$

$\bf\huge => 11\times 151 = 1661$

$\bf\huge Sum\:of\: first\:22\:term\:is\:1661$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$