Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{An}\hspace{0.17em}\mathrm{A}.\mathrm{P}.\mathrm{with}a=-6\mathrm{and}d=-\frac{11}{2}-\left(-6\right)=\frac{1}{2} \\ {S}_n=-25 \\ \therefore -25=\frac{n}{2}\left[2\times \left(-6\right)+\left(n-1\right)\frac{1}{2}\right] \\ \Rightarrow -25=\frac{n}{2}\left[-12+\frac{n}{2}-\frac{1}{2}\right] \\ \Rightarrow -50=n\left[\frac{n}{2}-\frac{25}{2}\right] \\ \Rightarrow -100=n\left(n-25\right) \\ \Rightarrow n^2-25n+100=0 \\ \Rightarrow \left(n-20\right)\left(n-5\right)=0 \\ \Rightarrow n=20\mathrm{or}n=5 \end{gathered}$$