Mathematics - Sequence & Series Question with Solution | TestHub

Let the A.P. be $\begin{array}{rl}& a,a+2,a+2d,\dots ,a+(2k-1)d\\ & \text{ Now, }a+a+2d+a+4d+\dots +a+(2k-2)d=40\\ & ka+2d+4d+\dots +(2k-2)d=40\\ & \Rightarrow ka+\frac{k-1}{2}[2d+2kd-2d]=40\end{array}$ $\Rightarrow ka+k(k-1)d=40$ ...(1) And $a+d+a+3d+\dots +a+(2k-1)d=55$ $\Rightarrow ka+\frac{k}{2}(d+2kd-d)=55$ $\Rightarrow ka+k^{2}d=55$ ....(2) Also, $a+(2k-1)d-a=27$ $\Rightarrow (2k-1)d=27\Rightarrow d=\frac{27}{2k-1}$ ...(3) From equation (1) and (2) $k^{2}d-kd-k^{2}d=-15$ $\Rightarrow d=\frac{15}{k}$ From equation (3) and (4) $\begin{array}{rl}& \frac{27}{2k-1}=\frac{15}{k}\\ & 27k=30k-15\\ & \Rightarrow 3k=15\\ & \Rightarrow k=5\end{array}$