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

**Given:** The sum of first n terms of an A.P. =$3n^{2} +4n$

**To do:** To find the $25^{th}$ term of the given A.P.

**Solution:**

The sum of first term $S_{n} =3n^{2} +4n$

For $n=1,\ S_{1} =3\times 1^{2} +4\times 1=3+4=7$

Thus its first term is , $a=7$

For $n=2,\ S_{2} =3\times 2^{2} \ +4\times 2=12+8=20$

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

$=20-7$

$=13$

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

$=13-7=6$

$\therefore \ 25^{th} \ term\ of\ the\ A.P.=a+( n-1) d$

$=7+( 25-1) \times 6$

$=7+24\times 6$

$=7+144$

$=151$

Therefore, The $25^{th}$ term of the given A.P. is 151.

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