Given  1/(x+2) , 1/(x+3), 1/(x+5) are in A.P.

So,  2/(x+3) = (1/x+2) + (1/x+5)

=>  2/(x+3) = {(x+2) + (x+5)}/{(x+2) * (x+5)}

=>  2/(x+3) = (2x+ 7)/( x² + 7x + 10)

=> 2*( x² + 7x + 10) = (x+3) * (2x+ 7)

=> 2*( x² + 7x + 10) = x² + 7x + 6x + 14

=> 2x² + 14x + 20 = 2x² + 13x + 21

=>  14x + 20 =  13x + 21

=>  14x -13x =  21 - 20

=>  x =1

So, the value of x =1