Question

The sum of first $7$ terms of an AP is $182$. If its $4^{th}$ and the ${17}^{th}$ terms are in the ratio $1:5$, find the AP.

Answer 1

$\begin{array}{c}{S}_7=182\\ \frac{T4}{T_{17}}=1:5,Ap=?let\text{a be a first term}andcommondifference\left(d\right)\\ S_7=\frac{7}{2}[2a+(7-1\left)d\right]\\ 182=\frac{7}{2}[2a+6d]\\ 182=\frac{7}{2}\times 2(a+3d)\\ 26=(a+3d)\\ 26=a+3d--\left(i\right)\\ \frac{T4}{T_{17}}=\frac{a+(4-1)d}{a+(17-1)d}=\frac{1}{5}\\ \frac{a+3d}{a+16d}=\frac{1}{5}\\ 5a+15d=a+16d\\ 4a=d.............\left(ii\right)\\ puttingd=4aine{q}^n\left(i\right)\\ 26=a+3\times 4a\\ 26=a+12a\\ 26=13a\\ a=2\\ andd=4a\\ d=4\times 2=8\\ Then,A.Pis\\ 2,10,18,26\end{array}$