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Question

Let $$T_{r}$$ be the $$r^{th}$$ term of an A.P., for $$r = 1, 2, ...$$ If for some positive integers $$m$$ and $$n, T_{m} = \dfrac {1}{n}; T_{n} = \dfrac {1}{m}$$ then find $$T_{mn}$$.


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 D $$1$$
$${ T }_{ m }=\cfrac { 1 }{ n } $$ (given)
$$a+\left( m-1 \right) d=\cfrac { 1 }{ n } \longrightarrow 1$$
$${ T }_{ n }=\cfrac { 1 }{ m } $$ (given)
$$a+\left( n-1 \right) d=\cfrac { 1 }{ m } \longrightarrow 2$$
Substract equation 2 from 1, we get
$$\left( m-1 \right) d-\left( n-1 \right) d=\cfrac { 1 }{ n } -\cfrac { 1 }{ m } $$
$$d\left[ m-1-n+1 \right] =\cfrac { m-n }{ mn } $$
$$d\left[ m-n \right] =\cfrac { m-n }{ mn } $$
$$d=\cfrac { 1 }{ mn } \longrightarrow 3$$
Put value of d in equation 1
$$a+\left( m-1 \right) \cfrac { 1 }{ mn } =\cfrac { 1 }{ n } $$
$$\cfrac { amn+\left( m-1 \right)  }{ mn } =\cfrac { 1 }{ n } $$
$$amn+m-1=m$$
$$a=\cfrac { 1 }{ mn } \longrightarrow 4$$
$${ T }_{ mn }=a+\left( mn-1 \right) d$$
$${ T }_{ mn }=\cfrac { 1 }{ mn } +\left( mn-1 \right) \cfrac { 1 }{ mn } $$ (From equation 3 & 4)
$${ T }_{ mn }=\cfrac { 1+mn-1 }{ mn } =\cfrac { mn }{ mn } $$
$${ T }_{ mn }=1$$

Mathematics

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