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