If 1x + 2, 1x + 3, 1x + 5 are in A.P. then x = ?

Answer: Option C

$$\begin{array}{rl}& \frac{1}{x+2},\frac{1}{x+3},\frac{1}{x+5}\text{are}\text{in}\text{A}\text{.P}\text{.}\\ & \therefore \frac{1}{x+3}-\frac{1}{x+2}=\frac{1}{x+5}-\frac{1}{x+3}\\ & \Rightarrow \frac{x+2-x-3}{\left(x+3\right)\left(x+2\right)}=\frac{x+3-x-5}{\left(x+5\right)\left(x+3\right)}\\ & \Rightarrow \frac{-1}{\left(x+3\right)\left(x+2\right)}=\frac{-2}{\left(x+5\right)\left(x+3\right)}\\ & \Rightarrow \frac{-1}{x+2}=\frac{-2}{x+5}\\ & \Rightarrow -2x-4=-x-5\\ & \Rightarrow -2x+x=-5+4\\ & \Rightarrow -x=-1\\ & \therefore x=1\end{array}$$