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Question

Let $$T_r$$ be the $$r^{th}$$ term of an A.P. for $$r = 1, 2, 3, ...$$ If for some positive integers $$m,\ n$$ we have $$T_m=\dfrac {1}{n}$$  and  $$ T_n = \dfrac {1}{m}$$,  then $$T_{mn}$$ 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,
$$T_m = a + (m - 1)d = \dfrac 1n\quad...(i)$$  and
$$T_n = a + (n - 1)d = \dfrac 1m\quad...(ii)$$

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

$$\{a+(m-1)d\}-\{1+(n-1)d\}=\dfrac 1n-\dfrac 1m\\ \Rightarrow a+md-d-a-nd+d=\dfrac{m-n}{mn}\\ \Rightarrow  (m - n)d =\dfrac{ (m - n)}{ mn}\\ \\ \Rightarrow \ \ d = \dfrac1{ mn}$$

Substituting the value of $$d$$ in equation $$(i)$$, we get
$$a+\dfrac{(m-1)}{mn}=\dfrac 1n\\ \Rightarrow a+\dfrac 1n-\dfrac 1{mn}=\dfrac 1n\\ \Rightarrow a-\dfrac 1{mn}=0\\ \Rightarrow a=\dfrac 1{mn}$$

$$\therefore  T_{mn} = a + (mn - 1)d \\ \qquad= \dfrac{1}{mn} + ( mn - 1) \dfrac{1}{mn}\\ \qquad= \dfrac{1}{mn} + 1 - \dfrac{1}{mn}=1$$. 

Hence, option (C) is the correct answer.

Maths

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