The sum of first n terms of an A.P.

Question:

The sum of first $n$ terms of an A.P. is $3 n^{2}+4 n$. Find the 25 th term of this A.P.

Solution:

$S_{n}=3 n^{2}+4 n$

We know

$a_{n}=S_{n}-S_{n-1}$

$\therefore a_{n}=3 n^{2}+4 n-3(n-1)^{2}-4(n-1)$

$\Rightarrow a_{n}=6 n+1$

$a_{25}=6(25)+1=151$

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