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

Solution:

Here, we are given three terms,

First term $\left(a_{1}\right)=\frac{1}{x+2}$

Second term $\left(a_{2}\right)=\frac{1}{x+3}$

Third term $\left(a_{3}\right)=\frac{1}{x+5}$

We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

$d=a_{2}-a_{1}$

$d=\left(\frac{1}{x+3}\right)-\left(\frac{1}{x+2}\right)$

$d=\frac{(x+2)-(x+3)}{(x+2)(x+3)}$

$d=\frac{x+2-x-3}{(x+2)(x+3)}$

$d=\frac{-1}{(x+2)(x+3)}$........(1)

Also,

$d=a_{3}-a_{2}$

$d=\left(\frac{1}{x+5}\right)-\left(\frac{1}{x+3}\right)$

$d=\frac{(x+3)-(x+5)}{(x+5)(x+3)}$

$d=\frac{x+3-x-5}{(x+5)(x+3)}$

$d=\frac{-2}{(x+5)(x+3)}$...........(2)

Now, on equating (1) and (2), we get,

$\frac{-2}{(x+5)(x+3)}=\frac{-1}{(x+3)(x+2)}$

$2(x+3)(x+2)=1(x+5)(x+3)$

$2 x+4=x+5$

$2 x-x=5-4$

$x=1$

Therefore, for $x=1$, these three terms will form an A.P.

So, the correct option is (c).