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The sum of the first n terms of an A.P. is $3n^2 + 6n$. Find the nth term of this A.P.
Given:
The sum of first $n$ terms of an A.P. is $3n^{2} +6n$.
To do:
We have to find the $n^{th}$ term of the given A.P.
Solution:
$S_{n} =3n^{2} +6n$
For $n=1,\ S_{1} =3\times 1^{2} +6\times 1=3+6=9$
Therefore, first term $a=9$
For $n=2,\ S_{2} =3\times 2^{2} \ +6\times 2=12+12=24$
$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$
$=24-9$
$=15$
Common difference of the A.P., $d=$second term $-$ first term
$=15-9=6$
We know that,
$a_{n}=a+(n-1)d$
$\therefore a_n=9+( n-1) \times 6$
$=9+6n-6$
$=6n+3$
Therefore, the $n^{th}$ term of the given A.P. is $6n+3$.
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