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Question

Let Tr be the rth term of an A.P. for r=1,2,3,... If for some positive integers m, n we have Tm=1n and Tn=1m, then Tmn equals

A
1mn
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B
1m+1n
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C
1
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D
0
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Solution

The correct option is D 1
Let a be the first term and d be the common difference of the A.P.
Then,
Tm=a+(m1)d=1n...(i) and
Tn=a+(n1)d=1m...(ii)

Subtracting equation (ii) from (i), we get

{a+(m1)d}{1+(n1)d}=1n1ma+mddand+d=mnmn(mn)d=(mn)mn d=1mn

Substituting the value of d in equation (i), we get
a+(m1)mn=1na+1n1mn=1na1mn=0a=1mn

Tmn=a+(mn1)d=1mn+(mn1)1mn=1mn+11mn=1.

Hence, option (C) is the correct answer.

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