Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

Solution:

Here, we are given three terms,

First term $\left(a_{1}\right)=2 x$

Second term $\left(a_{2}\right)=x+10$

 

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

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,

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

$d=(x+10)-(2 x)$

$d=x+10-2 x$

 

$d=10-x$ $\ldots(1)$

Also,

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

$d=(3 x+2)-(x+10)$

$d=3 x+2-x-10$

 

$d=2 x-8$$\ldots(2)$

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

$10-x=2 x-8$

 

$2 x+x=10+8$

$3 x=18$

$x=\frac{18}{3}$

 

$x=6$

Therefore, for $x=6$, these three terms will form an A.P.