n + 1
2n + 1
3n + 1
4n + 1
Correct answer is B
Sn=n2(2a+(n−1)d=n2+2n
n(2a+(n−1)d=2n2+4n
2an+n2d−nd=2n2+4n
n2d=2n2
d=2
(2a−d)n=4n
2a−d=4⟹2a=4+d=4+2=6
a=3
Tn=a+(n−1)d
= 3+(n−1)2=3+2n−2=2n+1
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