8. Sequences and Series
hard

Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is

A

$n + 4n^2$

B

$6n^2 - n$

C

$n^2 + 4n$

D

$3n + 2n^2$

(JEE MAIN-2013)

Solution

Given ${S_n} = 2n + 3{n^2}$

Now, first term $=2+3=5$

second term $2(2)+3(4)=16$

third term $=2(3)+3(9)=33$

Now, sum given in option $(b)$ only has the same first term and difference between ${2^{nd}}$ and ${1^{st}}$ term is double also.

Standard 11
Mathematics

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