Question

Question 24
In an AP, if $S_n$ = n(4n+1), then find the AP.

Answer 1

We know that, the nth term of an AP is;
$a_n=S_n-S_{n-1}$
$a_n=n(4n+1)-(n-1)\left\{4\right(n-1)+1\}[\because S_n=n(4n+1\left)\right]$
$\Rightarrow a_n=4n^2+n-(n-1)(4n-3)$
$\Rightarrow a_n=4n^2+n-4n^2+3n+4n-3$
$\Rightarrow a_n=8n-3$

$Putn=1,\Rightarrow a_1=8\left(1\right)-3=5$
$Putn=2,\Rightarrow a_2=8\left(2\right)-3=16-3=13$
$Putn=3,\Rightarrow a_3=8\left(3\right)-3=24-3=21$

Hence, the required AP is 5, 13, 21, . . .