Suppose that the number of terms in an A.P. is 2 k, k ∈ N . If the...

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