The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

Answer: Option B

First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be the number of terms and d be the common difference, then
$$\begin{array}{rl}& a_n=1=a+\left(n-1\right)d=11\\ & \Rightarrow 1+\left(n-1\right)d=11\\ & \Rightarrow \left(n-1\right)d=11-1\\ & \Rightarrow \left(n-1\right)d=10.....\left(1\right)\\ & S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=36\end{array}$$
$$\Rightarrow \frac{n}{2}\left[2\times 1+10\right]=36$$       $$\left[\text{From}\left(1\right)\right]$$
$$\begin{array}{rl}& \Rightarrow n\left(2+10\right)=72\\ & \Rightarrow 12n=72\\ & \Rightarrow n=\frac{72}{12}\\ & \Rightarrow n=6\end{array}$$