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