Question

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

Answer 1

Given $A.P.$ is $-6,-\frac{11}{2},-5,\dots$
$a=-6$
$d=-\frac{11}{2}-(-6)=\frac{1}{2},S_n=-25$
$\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]=-25$
$\Rightarrow \frac{n}{2}\left[2\right(-6)+(n-1\left)\left(\frac{1}{2}\right)\right]=-25$
$\Rightarrow n[-12+(n-1\left)\left(\frac{1}{2}\right)\right]=-50$
$\Rightarrow n[-24+(n-1\left)\right]=-100$
$\Rightarrow n(n-25)=-100$
$\Rightarrow n^2-25n+100=0$
$\Rightarrow (n-20)(n-5)=0$
$\Rightarrow n=5,20$
Therefore, the required number of terms is $20$ or $5$.