Question

How many terms of the AP 3, 7, 11, 15, ... will make the sum 406?
(a) 10
(b) 12
(c) 14
(d) 20

Answer 1

(c) 14

Here, a = 3 and d = (7-3) = 4
Let the sum of n terms be 406 .
Then, we have:
$$\begin{gathered} Sn=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=406 \\ \Rightarrow \frac{n}{2}\left[2\times 3+\left(n-1\right)\times 4\right]=406 \\ \Rightarrow n\left[3+2n-2\right]=406 \\ \Rightarrow 2n^2+n-406=0 \\ \Rightarrow 2n^2-28n+29n-406=0 \\ \Rightarrow 2n(n-14)+29(n-14)=0 \\ \Rightarrow (2n+29)(n-14)=0 \\ \Rightarrow n=14(\because n\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{a}\mathrm{fraction}) \end{gathered}$$

Hence, 14 terms will make the sum 406.