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Q.

Given sum of the first n terns of an AP is 2n + 3 n2. Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is

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a

n+4n2

b

6n2−n

c

n2+4n

d

3n+2n2

answer is B.

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Detailed Solution

Here,T1=S1=2(1)+3(1)2=5T2=S2−S1=16−5=11∵S2=2(2)+3(2)2=16T3=S3−S2=33−16=17 ∵S3=2(3)+3(3)2=33Hence, sequence is 5, 11 , 17.∴ a=5 and d=6For new AP,A=5,D=2×6=12∴ Sn′=n2[2×5+(n−1)12]=6n2−n

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