Question

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

Answer 1

It is known that, $S_n=\frac{n}{20}[2a+(n-1\left)d\right]$,
where $n=$ number of terms , $a=$ first term, and $d=$ common difference
Here, $a=-6,d=\frac{1}{2}$
$\Rightarrow -25=\frac{n}{2}[2\times (-6)+(n-1\left)\left(\frac{1}{2}\right)\right]\Rightarrow -50=n[-12+\frac{n}{2}-\frac{1}{2}]\Rightarrow -50=n[-\frac{25}{2}+\frac{n}{2}]\Rightarrow -100=n(-25+n)\Rightarrow n^2-25n+100=0\Rightarrow n^2-5n-20n+100=0\Rightarrow n(n-5)-20(n-5)=0\Rightarrow n=20$

$\Rightarrow n=5$