If 4/5, a, 2 are three consecutive terms of an A.P., then find the value of a.

Here, we are given three consecutive terms of an A.P.

First term $\left(a_{1}\right)=\frac{4}{5}$

Second term $\left(a_{2}\right)=a$

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

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

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

$d=a-\frac{4}{5}$$\ldots(1)$

Also,

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

$d=2-a$$\cdots(2)$

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

$a-\frac{4}{5}=2-a$

$a+a=2+\frac{4}{5}$

$2 a=\frac{10+4}{5}$

$a=\frac{14}{10}$

$a=\frac{7}{5}$

Therefore, $a=\frac{7}{5}$