An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Given, an A.P of $60$ terms

And, $a=7$ and ${{a}_{60}}=125$

We know that ${{a}_{n}}=a+\left( n-1 \right)d$

⇒ ${{a}_{60}}=7+\left( 60-1 \right)d=125$

$7+59d=125$

$59d=118$

$d=2$

So, the ${{32}^{nd}}$ term is given by

${{a}_{32}}=7+\left( 32-1 \right)2=7+62=69$

⇒ ${{a}_{32}}=69$