If the sum of first $n$ terms of an A.P. is $\frac{1}{2}(3n^2 + 7n)$, then find its $n$th term. Hence write its 20th term.

Given:

The sum of first $n$ terms of an A.P. is $\frac{1}{2}(3n^2 + 7n)$.

To do:

We have to find the $n$th term and $20^{th}$ term of the given A.P.

Solution:

$S_{n} =5n^{2} +3n$

For $n=1,\ S_{1} =\frac{1}{2}[3(1)^2 + 7(1)]=\frac{1}{2}(10)=5$

Therefore, first term $a=5$

For $n=2,\ S_{2} =\frac{1}{2}[3(2)^2 + 7(2)]=\frac{1}{2}(26)=13$

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

$=13-5$

$=8$

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

$=8-5=3$

We know that,

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

$a_{n}=5+( n-1) \times 3$

$=5+3n-3$

$=3n+2$

$a_{20}=5+( 20-1) \times 3$

$=5+19\times 3$

$=5+57$

$=62$

Therefore, $n$th term is $3n+2$ and the $20^{th}$ term of the given A.P. is $62$.