Question

The sum of first 7 terms of an AP is 10, and that of the next 7 tems is 17, find the 8th term of the AP

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{d}\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{difference}. \\ \mathrm{Sum}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{terms}\mathrm{is}, \\ {\mathrm{S}}_n\mathit{}=\frac{n}{2}\left[2a+\left(n-1\right)\mathit{}d\right] \\ \Rightarrow \mathit{}{S}_{\mathit{7}}=\frac{7}{2}\left[2a+\left(7-1\right)d\right][\mathrm{here},\mathrm{n}=7] \\ \Rightarrow 10=\frac{7}{2}\left[2a\mathit{}+6d\right] \\ \Rightarrow 10=7\left[a+3d\right] \\ \Rightarrow 7a+21d=10............\left(1\right) \\ \mathrm{also},S_{14}=\frac{14}{2}\left[2a+(14-1)d\right]=7(2a+13d)[\mathrm{Here}\mathrm{n}=14] \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}: \\ {S}_{14}-\mathit{}{S}_7=17 \\ \Rightarrow 7\left[2a+13d\right]-10=17 \\ \Rightarrow 14a+91d=27..........\left(2\right) \\ \mathrm{Multiply}\left(1\right)\mathrm{by}\text{'}2\text{'},\mathrm{we}\mathrm{get} \\ 14a+42d=20........\left(3\right) \\ \mathrm{Subtracting}\left(3\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 49d=7 \\ \Rightarrow \mathit{}d\mathit{}=\frac{7}{49} \\ \Rightarrow \mathit{}d=\frac{1}{7} \\ \mathrm{Substituting}\mathit{}d=\frac{1}{7}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 7a+21\times \frac{1}{7}=10 \\ \Rightarrow 7a+3=10 \\ \Rightarrow 7a=10-3 \\ \Rightarrow 7a=7 \\ \Rightarrow \mathit{}a=1 \\ \mathrm{Therefore}a=1\mathrm{and}d\mathit{}=\frac{1}{7} \end{gathered}$$

${\mathrm{a}}_8=\mathrm{a}+7\mathrm{d}=1+7\times \frac{1}{7}=1+1=2$