Question

If $\frac{1}{x+2},\frac{1}{x+3}$ and $\frac{1}{x+5}$ are in A.P. Find the value of $x$.

Answer 1

Since $\frac{1}{x+2},\frac{1}{x+3},\frac{1}{x+5}$ are in AP

We can write the first term as $a_1=\frac{1}{x+2}$,

second term as $a_2=\frac{1}{x+3}$

and $a_3=\frac{1}{x+5}$

So, these terms are in A.P, then

$a_2-a_1=a_3-a_2$ [difference between consecutive terms will be equal.]

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

$\Rightarrow \frac{2}{x+3}=\frac{1}{x+2}+\frac{1}{x+5}$

$\Rightarrow \frac{2}{x+3}=\frac{2x+7}{(x+2)(x+5)}$

$\Rightarrow 2(x+2)(x+5)=(x+3)(2x+7)$

$\Rightarrow 2(x^2+7x+10)=2x^2+7x+6x+21$

$\Rightarrow 2x^2+14x+20=2x^2+13x+21$

$\Rightarrow 14x-13x=21-20$

$\therefore x=1$