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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