Find the value of x for which (8x + 4), (6x − 2) and

Solution:

Here, we are given three terms,

First term $\left(a_{1}\right)=8 x+4$

Second term $\left(a_{2}\right)=6 x-2$

Third term $\left(a_{3}\right)=2 x+7$

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,

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

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

$d=6 x-8 x-2-4$

$d=-2 x-6 \ldots \ldots(1)$

Also,

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

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

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

$d=-4 x+9 \ldots \ldots(2)$

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

$-2 x-6=-4 x+9$

$4 x-2 x=9+6$

$2 x=15$

$x=\frac{15}{2}$

Therefore, for $\left.x=\frac{15}{2}\right]$, these three terms will form an A.P.