Question

If 1/x+2,1/x+3and 1/x+5 are in AP. Find the value of x.

Answer 1

Heya !!!

This is your answer.

Given AP :-

$\frac{1}{x + 2}.. \frac{1}{x + 3} .. \frac{1}{x + 5} ......$
Here,
First term, a = 1/(X+2).

We know that a2 - a1 = a3 - a2. ......(I)

So, we will use this to find the value of x.

Let us find
a2-a1
$\frac{1}{x + 3} - \frac{1}{x + 1} \\ = \frac{(x + 1) - (x + 3)}{(x + 1)(x + 3)} \\ = > x + 1 - x - 3 = {x}^{2} + (1 + 3)x + 3 \\ = > - 2 = {x}^{2} + 4x + 3 \\ = > {x}^{2} + 4x + 5 = 0$
Hence, x² + 4x +5 = 0. ........(ii).

Now,
a3 - a2
$\frac{1}{x + 5} - \frac{1}{x + 3} \\ = > \frac{(x + 3) - (x + 5)}{(x + 5)(x + 3)} \\ = > x + 3 - x - 5 = {x}^{2} + (5 + 3)x + 3 \times 5 \\ = > - 2= {x}^{2} + 8x + 15 \\ = > {x}^{2} + 8x + 17 = 0.$
Hence, x² +8x + 17 = 0. .......(iii).

From equation (i), (ii) and (iii), we get
x² +4x + 5 = x² + 8x + 17
=> 4x + 5 = 8x + 17
=> 8x - 4x = 5 - 17
=> 4x = -12
=> x = -12/4
x = -3.

Hence, the value of x is -3.

Hope you got the answer.