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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 $$m,n$$ we have $${T}_{m}=1/n$$ and  $${T}_{n}=1/m$$, then $${T}_{m/n}$$ equal to ________


A
1mn
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B
1m+1n
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C
1n2
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D
0
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Solution

The correct option is A $$\cfrac{1}{n^2}$$
We  have
$${T_m} = \frac{1}{n}$$ and$${T_n} = \frac{1}{m}$$  

$${T_m} = \frac{1}{n} = a + \left( {n - 1} \right)d\,\,\,\,\,\,\,\,\,\, -  -  - \left( 1 \right)$$
And
$${T_n} = \frac{1}{m} = a + \left( {n - 1} \right)d\,\,\,\,\,\,\,\,\,\, -  -  - \left( 2 \right)$$

By $$(1)-(2)\,\,equ^n$$ We get,

$${T_m} - {T_n} = \frac{1}{n} - \frac{1}{m} = \left( {m - n} \right)d$$

$$d = \frac{1}{{mn}}\,\,\,\,\,\,\,\,\,\, -  -  - \left( 3 \right)$$
By putting value of $$d$$ in $$equ^n \,\,(1)$$ we get,
$$a = \frac{1}{n} - \frac{{\left( {m - 1} \right)}}{{mn}}$$

$${T_{\frac{m}{n}}} = \frac{1}{n} - \frac{{m - 1}}{{mn}} + \frac{1}{{mn}}.\left( {\frac{m}{n} - 1} \right)$$

$$\,\,\,\,\, = \frac{1}{n} - \frac{1}{n} + \frac{1}{{mn}} + \frac{1}{{{n^2}}} - \frac{1}{{mn}}$$ 
  
$$\,\,\,\,\,\, = \frac{1}{{{n^2}}}$$

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