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
B
1m+1n
C
1
D
0

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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