Question

How many terms of the AP −15, −11, −7, ... are needed to make the sum 744?

Answer 1

Here, a = −15 and d = [−11 − (−15)] = 4

Let the required number of terms be n. Then we have:

$$\begin{gathered} S_n=744 \\ \Rightarrow \frac{n}{2}\left[2a+\left(n-1\right)d\right]=744 \\ \Rightarrow \frac{n}{2}\left[2\times (-15)+\left(n-1\right)\times 4\right]=744 \\ \Rightarrow \frac{n}{2}\left[-30+4n-4\right]=744 \\ \Rightarrow n[2n-17]=744 \\ \Rightarrow 2n^2-17n-744=0 \\ \Rightarrow 2n^2-48n+31n-744=0 \\ \Rightarrow 2n(n-24)+31(n-24)=0 \\ \Rightarrow (2n+31)(n-24)=0 \\ \Rightarrow n=24orn=-31/2(n=\frac{-31}{2}\mathrm{rejected},\mathrm{as}n\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{a}\mathrm{fraction}) \end{gathered}$$
Hence, 24 terms are needed to make the sum of 744.