How many terms of the A.P.

Question:

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

Solution:

Given:

An A.P. with $a=-6$ and $d=-\frac{11}{2}-(-6)=\frac{1}{2}$

$S_{n}=-25$

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

$\Rightarrow-25=\frac{n}{2}\left[-12+\frac{n}{2}-\frac{1}{2}\right]$

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

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

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

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

$\Rightarrow n=20$ or $n=5$

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