Question

if one by X + 2, 1 by X + 3 and 1 by X + 5 are in AP find the value of x

Answer 1

Answer:

1

Step-by-step explanation:

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

Now,

We know,

In an A.P. the common difference is equal for all the terms

So,

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

and,

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

i.e.  

$a_{2} - a_{1}$ = $a_{3} - a_{2}$

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

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

We can cancel (x + 3) from both sides,

then,

=> - 1 / (x+2) = - 2 / (x+5)

=> - x - 5 = - 2x - 4

=> x = 5 - 4

=> x = 1