Question

If $\frac{3+5+7+...+\mathrm{upto}n\mathrm{terms}}{5+8+11+....\mathrm{upto}10\mathrm{terms}}$ 7, then find the value of n.

Answer 1

$$\begin{gathered} \frac{3+5+7+...+\mathrm{upto}n\mathrm{terms}}{5+8+11+....\mathrm{upto}10\mathrm{terms}}=7 \\ \Rightarrow \frac{{\displaystyle \frac{n}{2}}\left\{2\times 3+\left(n-1\right)2\right\}}{{\displaystyle \frac{10}{2}}\left\{2\times 5+\left(10-1\right)3\right\}}=7 \\ \Rightarrow \frac{n\left(4+2n\right)}{370}=7 \\ \Rightarrow n^2+2n-1295=0 \\ \Rightarrow n^2+37n-35n-1295=0 \\ \Rightarrow \left(n+37\right)\left(n-35\right) \\ \Rightarrow n=35,n=-37 \\ \mathrm{Rejecting}\mathrm{the}\mathrm{negative}\mathrm{value},\mathrm{we}\mathrm{get}: \\ \Rightarrow n=35 \end{gathered}$$