In an \( \mathrm{AP} \), if \( \mathrm{S}_{n}=n(4 n+1) \), find the \( \mathrm{AP} \).

Given:

In an \( \mathrm{AP} \),  \( \mathrm{S}_{n}=n(4 n+1) \)

To do:

We have to find the \( \mathrm{AP} \).

Solution:

 Substituting $n=1$, we get,

$\mathrm{S}_{1}=1(4 \times1+1)$

$=4+1$

$=5$

This implies,

$a_1=a=5$

$\mathrm{S}_{2}=2(4 \times2+1)$

$=2(8+1)$

$=18$

$S_2=a_1+a_2$

$18=a+a+d$

$18=2a+d$

$18=2(5)+d$

$d=18-10$

$d=8$

Therefore,

$a_2=a+d$

$=5+8$

$=13$

$a_3=a+2d$

$=5+2(8)$

$=5+16$

$=21$

The required AP is $5,13,21,......$

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Updated on: 10-Oct-2022

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