Answer: Option C
$$\eqalign{
& \frac{1}{{x + 2}},\,\frac{1}{{x + 3}},\,\frac{1}{{x + 5}}\,{\text{are}}\,{\text{in}}\,{\text{A}}{\text{.P}}{\text{.}} \cr
& \therefore \frac{1}{{x + 3}} - \frac{1}{{x + 2}} = \,\frac{1}{{x + 5}} - \frac{1}{{x + 3}} \cr
& \Rightarrow \frac{{x + 2 - x - 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{x + 3 - x - 5}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr
& \Rightarrow \frac{{ - 1}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr
& \Rightarrow \frac{{ - 1}}{{x + 2}} = \frac{{ - 2}}{{x + 5}} \cr
& \Rightarrow - 2x - 4 = - x - 5 \cr
& \Rightarrow - 2x + x = - 5 + 4 \cr
& \Rightarrow - x = - 1 \cr
& \therefore x = 1 \cr} $$