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Correct option is
D
9
We know that sum of n terms of an A.P. is given by
Sk=2k{2a1+(k−1)d}
So,SnSm=SnS5n=2n{6+(n−1)d}25n{6+(5n−1)d}=n{6+(n−1)d}5n{6+(5n−1)d}={(6−d)+n}5{(6−d)+5nd}
Since, SnSm is independent of n , so d=6.
If d=6 then a2=a1+d=3+6=9
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