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 $62$.
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