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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
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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 C $$1$$
$$T_m= a+(m-1) d=\dfrac1n$$

$$T_n=a+(n-1)d=\dfrac1m$$

$$\implies d= \dfrac1{mn}, a=\dfrac1m-\dfrac{n-1}{mn}$$
$$T_{mn} = a+(mn-1)d =\dfrac1m-\dfrac1m+\dfrac1{mn} +\dfrac{mn-1}{mn}= 1$$

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