Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?

In the given problem, let us first find the 13th term of the given A.P.

A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. $(d)=10-3=7$

Now, as we know,

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

So, for 13th term (n = 13),

$a_{13}=3+(13-1)(7)$

$=3+12(7)$

$=3+84$

$=87$

Let us take the term which is 84 more than the 13th term as an. So,

$a_{n}=84+a_{13}$

$=84+87$

$=171$

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

$171=3+(n-1) 7$

$171=3+7 n-7$

$171=-4+7 n$

$171+4=7 n$

Further simplifying, we get,

$175=7 n$

$n=\frac{175}{7}$

$n=25$

Therefore, the $25^{\text {th }}$ term of the given A.P. is 84 more than the $13^{\text {th }}$ term.