\begin{aligned} & a_1, a_2, a_3, \ldots \ldots, a_{2 k} \rightarrow \text { A.P. } \\ & \sum_{r=1}^k a_{2 r-1}=40, \sum_{r=1}^k a_{2 r}=55, a_{2 k}-a_1=27 \\ & \frac{k}{2}\left[2 a_1+(k-1) 2 d\right]=40, \frac{k}{2}\left[2 a_2+(k-1) 2 d\right]=55, \\ & d=\frac{27}{2 k-1} \\ & a_1=\frac{40}{k}-(k-1) d=\frac{55}{k}-k d \\ & d=\frac{15}{k} \Rightarrow \frac{27}{2 k-1}=\frac{15}{k} \Rightarrow 9 k=10 k-5 \\ & \therefore k=5 \end{aligned}