Question

# Let $$T_r$$ be the $$r$$th term of an AP for $$r=1, 2...$$. If for some positive integers $$m$$ and $$n$$ we have $$T_m= \dfrac1n$$ and $$T_n=\dfrac1m$$, then $$T_{mn} =$$

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

Solution

## The correct option is C $$1$$$$T_m= a+(m-1) d=\dfrac1n$$$$T_n=a+(n-1)d=\dfrac1m$$$$\implies d= \dfrac1{mn}, a=\dfrac1m-\dfrac{n-1}{mn}$$$$T_{mn} = a+(mn-1)d =\dfrac1m-\dfrac1m+\dfrac1{mn} +\dfrac{mn-1}{mn}= 1$$Maths

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