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Question

For an A.P if SmSn=m2n2, then tmtn=______

A
2n12m+1
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B
2m12n1
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C
mn
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D
nm
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Solution

The correct option is D 2m12n1
Let a be the first term and d be the common difference of the given A.P
Now,
SmSn=m2n2

m2|2a+(m1)d|n2|2a+(n1)d|=m2n2 [ cancle 2 both numerator and denominator and also cancle m and n both sides]

2a+(m1)d2a+(n1)d=mn(i)
Now,
tmtn=a+(m1)da+(n1)d

=2a+2(m1)d2a+2(n1)d

=2a+(2m2)d2a+(2n2)d

tmtn==2a+[2(m1)d]2a+[2(n1)d](ii)
Put 2m1 in place of p and (2n1) in place of q in (i) we get,
2a+[2(m1)d]2a+[2(n1)d]=2m12n1(iii)

From (ii) and (iii) we have
tmtn=2m12n1
Hence, the option (B) is correct.


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