An A.P. consists of 60 terms.

Solution:

In the given problem, we need to find the 32nd term of an A.P. which contains a total of 60 terms.

Here we are given the following,

First term (a) = 7

Last term (an) = 125

Number of terms (n) = 60

So, let us take the common difference as d

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$125=7+(60-1) d$

 

$125=7+(59) d$

$125-7=59 d$

$118=59 d$

Further simplifying,

$d=\frac{118}{59}$

$d=2$

So, for the 32nd term (n = 32)

$a_{32}=7+(32-1) 2$

$=7+(31) 2$

$=7+62$

$=69$

Therefore, the $32^{\text {nd }}$ term of the given A.P. is 69 .