Question

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

Answer 1

Here a = -6

$=\frac{-11}{2}-(-6)=\frac{-11}{2}+6=\frac{1}{2}.$

$s_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

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

$\Rightarrow -50=n\left[\frac{-24+n-1}{2}\right]$

$\Rightarrow -100=n^2-25n$

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

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

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

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

$Eithern-20=0orn-5=0$

$\Rightarrow n=20orn=5$