Question

How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=-14\mathrm{and}{S}_n=40...\left(\mathrm{i}\right) \\ {a}_5=2 \\ \Rightarrow a+\left(5-1\right)d=2 \\ \Rightarrow -14+4d=2 \\ \Rightarrow 4d=16 \\ \Rightarrow d=4...\left(\mathrm{ii}\right) \\ \mathrm{Also},S_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow 40=\frac{n}{2}\left[2\left(-14\right)+(n-1)\times 4\right]\left(\mathrm{From}\right(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right) \\ \Rightarrow 80=n\left[-28+4n-4\right] \\ \Rightarrow 80=4n^2-32n \\ \Rightarrow n^2-8n-20=0 \\ \Rightarrow (n-10)(n+2)=0 \\ \Rightarrow n=10,-2 \\ \mathrm{But},\mathrm{n}\mathrm{cannot}\mathrm{be}\mathrm{negative}. \\ \therefore \mathrm{n}=10 \end{gathered}$$