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Question
Sn=2n2+3n; then d=..........
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Solution
The correct option is B 4
Given : Sn=2n2+3nnth term of any series is given by
Tn=Sn−Sn−1
=(2n2+3n)−[2(n−1)2+3(n−1))
=2n2+3n−[2n2−4n+2+3n−3)
∴Tn=4n+1
Now d= common difference
∴ d=T2−T1=[4(2)+1]−[4(1)+1]=9−5=4Hence, d=4
Given : Sn=2n2+3n
nth term of any series is given by
Tn=Sn−Sn−1
=(2n2+3n)−[2(n−1)2+3(n−1))
=2n2+3n−[2n2−4n+2+3n−3)
∴Tn=4n+1
Now d= common difference
∴ d=T2−T1=[4(2)+1]−[4(1)+1]=9−5=4
=(2n2+3n)−[2(n−1)2+3(n−1))
=2n2+3n−[2n2−4n+2+3n−3)
∴Tn=4n+1
Now d= common difference
∴ d=T2−T1=[4(2)+1]−[4(1)+1]=9−5=4
Hence, d=4
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