If the sum of first $n$ terms of an A.P. is $n^2$, then find its 10th term.

Given:

The sum of the first $n$ terms of an A.P. is $n^2$.

To do:

We have to find the $10^{th}$ term of the given A.P.

Solution:

$S_{n} =n^2$

For $n=1,\ S_{1} =(1)^2=1$

Therefore, first term $a=1$

For $n=2,\ S_{2} =(2)^2=4$

$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$

$=4-1$

$=3$

Common difference of the A.P., $d=$second term $-$ first term

$=3-1=2$

We know that,

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

$\therefore a_{10}=1+( 10-1) \times 2$

$=1+9\times 2$

$=1+18$

$=19$

Therefore, the $10^{th}$ term of the given A.P. is $19$.