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Question

The sum of 'r' terms of an AP is (2r2+3r). The nth​ term is :


A

2n - 1

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B

4n + 1

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C

3n - 1

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D

3n + 1

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Solution

The correct option is B

4n + 1


Sr=(2r2+3r)
Tn=SnSn1
So, Sn=2n2+3n
Sn1=2(n1)2+3(n1)
Now,
Tn=SnSn1
=2n2+3n[2(n22n+1)+3n3]
=2n2+3n2n2+4n23n+3=4n+1

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