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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 B $$1$$

Let, common difference$$=d$$

$${{r}^{th}}$$ term $$={{t}_{r}}$$

$${{1}^{st}}$$ term= a

Now $${{m}^{th}}$$ term= $${{t}_{m}}=a+\left( m-1 \right)d$$

Since, $${{t}_{m}}=\dfrac{1}{n}$$

Therefore,

$$\dfrac{1}{n}=a+\left( m-1 \right)$$      …… (1)

Now $${{n}^{th}}$$ term $${{t}_{n}}=a+\left( m-1 \right)d$$

Given ,$${{t}_{n}}=\dfrac{1}{m}$$

Therefore,

$$\dfrac{1}{m}=a+\left( n-1 \right)$$      …… (2)

Subtract equation $$\left( 2 \right)$$ from equation$$\left( 1 \right)$$

  $$ \dfrac{1}{n}-\dfrac{1}{m}=\left( m-1 \right)d-\left( n-1 \right)d $$

 $$ \dfrac{m-n}{mn}=\left( m-n \right)d $$

 $$ d=\dfrac{1}{mn} $$

Put value d in equation $$\left( 1 \right)$$

  $$ \dfrac{1}{n}=a+\left( m-1 \right)\dfrac{1}{mn} $$

 $$ \dfrac{1}{n}=a+\dfrac{m}{mn}+\dfrac{1}{mn} $$

 $$ \dfrac{1}{n}=a+\dfrac{1}{n}-\dfrac{1}{mn} $$

 $$ \dfrac{1}{n}=a+\dfrac{1}{n}-\dfrac{1}{mn} $$

 $$ a=\dfrac{1}{mn} $$

Now $$m{{n}^{th}}$$ term ,

  $$ {{t}_{mn}}=\dfrac{1}{mn}+\left( mn-1 \right)\dfrac{1}{mn} $$

 $$ =\dfrac{1}{mn}+1-\dfrac{1}{mn} $$

 $$ {{t}_{mn}}=1 $$


Mathematics

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