# How many terms of the A.P. are needed to give the sum –25?

Question:

How many terms of the A.P. $-6,-\frac{11}{2},-5, \ldots$ are needed to give the sum $-25 ?$

Solution:

Let the sum of n terms of the given A.P. be –25.

It is known that, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, where $n=$ number of terms, $a=$ first term, and $d=$ common difference

Here, $a=-6$

$d=-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}$

Therefore, we obtain

$-25=\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]$

$\Rightarrow-50=n\left[-12+\frac{n}{2}-\frac{1}{2}\right]$

$\Rightarrow-50=n\left[-\frac{25}{2}+\frac{n}{2}\right]$

$\Rightarrow-100=n(-25+n)$

$\Rightarrow n^{2}-25 n+100=0$

$\Rightarrow n^{2}-5 n-20 n+100=0$

$\Rightarrow n(n-5)-20(n-5)=0$

$\Rightarrow n=20$ or 5