If x + 1, 3x and 4x + 2 are in A.P., find the value of x.

Solution:

Here, we are given three terms which are in A.P.,

First term (a1) = 

Second term (a2) = 

Third term (a3) = 

We need to find the value of x. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

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

$d=(3 x)-(x+1)$

$d=3 x-x-1$

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

Also,

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

$d=(4 x+2)-(3 x)$

$d=4 x-3 x+2$

$d=x+2 \ldots \ldots(2)$

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

$2 x-1=x+2$

 

$2 x-x=2+1$

$x=3$

Therefore, for $x=3$, these three terms will form an A.P.