Question

The sum of how many terms of an A.P. : $7,11,15,19,23,$ ..... will be $900$?

Answer 1

The sum is $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

where

$S_n=$Sum of an AP

$a=$first term of an AP

$n=$number of terms of an AP

$d=$difference of an AP

In the question: $a=7,d=4,S_n=900$

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$900=\frac{n}{2}(14+(n-1\left)4\right)$

$900=\frac{n}{2}(10+4n)$

$900=\frac{n}{2}\left(2\right(5+2n\left)\right)$

$900=n(5+2n)$

$900=2n^2+5n$

$2n^2+5n-900=0$

$n=\frac{-5\pm \sqrt{25+4\times 2\times 900}}{4}$

$n=\frac{-5\pm \sqrt{7225}}{4}$

$n=\frac{-5\pm 85}{4}$

$n=-22.5,20$

$n$ cannot be negative hence $n=20$

hence the answer is $n=20$