Question

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

Options

• 5

• 3

• 1

• 2

Answer 1

Here, we are given three terms,

First term (a₁) = `1/(x + 2)`

Second term (a₂) =  `1/(x + 3)`

Third term (a₃) = `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 = a₂ - a₁

`d = (1/(x + 3)) - (1/(x + 2 ))`

`d = ((x + 2) - (x - 3))/((x + 2)(x + 3))`

`d =(x +2-x - 3)/((x + 2)(x + 3))`

`d =( -1) /((x + 2)(x +3 ))`                         ...............(1) 

Also,

`d = a_3 - a_2`

`d = (1/(x +5 ) ) - (1/(x + 3))`

`d = (( x + 3) - ( x + 5) ) /((x + 5)(x +3 ))`

`d = (x +3 - x - 5)/((x + 5)(x + 3))`

`d = (-2)/((x + 5)(x + 3))`                  .............(2) 

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

`(-2)/((x +5)(x + 3)) = (-1)/((x + 3)(x +2 ))`

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

             2x + 4 = x +5 

            2x - x  =  5 - 4

                    x = 1

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