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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.
Updated on: 10-Oct-2022
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