The sum of first $ n $ terms of an AP is given by $ \mathrm{S}_{n}=4 n^{2}+n $. Find that $ \mathrm{AP} $.
Given:
$S_n=4 n^{2}+n $
To do:
We have to find the AP.
Solution:
Let us take sum upto 1 term
$S_1=4 (1)^{2}+(1)=4+1=5$
Let us take sum upto two terms
$S_2=4 (2)^{2}+(2)=16+2=18$
We know,
$S_1=a_1=5$
$S_2=a_1+a_2=18$
$S_2-S_1=a_1+a_2-a_1$
$18-5=a_2$
$a_2=13$
We know that $d=a_2-a_1$
d=$13-5=8$
This implies,
$a_3=a_2+d=13+8=21$
$a_4=a_3+d=21+8=29$
The required AP is $5, 13, 21, 29,.....$
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