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


If $\frac{1}{x+2}, \frac{1}{x+3}, \frac{1}{x+5}$ are in A.P. Then, $x=$

(a) 5

(b) 3

(c) 1

(d) 2


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,












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



$2 x+4=x+5$

$2 x-x=5-4$


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

So, the correct option is (c).

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