Question

Find the value of x for which (8x + 4), (6x − 2) and (2x + 7) are in A.P.

Answer 1

Here, we are given three terms,

First-term `(a_1) = 8x + 4`

Second term `(a_2) = 6x - 2`

Third term `(a_3) = 2x + 7`

We need to find the value of x for which these terms are in A.P. So, in as A.P. the difference of two adjacent terms is always constant. So we get

`d = a_2 - a_1`

d = (6x - 2) - (8x + 4)

d = 6x - 8x -2 - 4

d = -2x - 6 ......(1)

Also

`d = a_3 - a_2`

d = (2x + 7) - (6x - 2)

`d = 2x - 6x + 7 + 2`

d = -4x + 9 ....(2)

Now on equating 1and 2 we get
`-2x - 6 = -4x  + 9`

4x - 2x = 9 + 6

2x = 15

`x = 15/2`

Therefore for `x = 15/2` these three terms will form an A.P