Arithmetically correct

Algebra Level 4

Let ${ T }_{ r }$ be the ${ r }^{\text{ th} }$ term of an arithmetic progression. for $r = 1,2,3, \ldots$ . If for some positive integers $m, n$ , we have ${ T }_{ m }=\frac { 1 }{ n }$ , ${ T }_{ n }=\frac { 1 }{ m }$ , then ${ T }_{ mn }$ equals to?

Practice for Bitsat here .

\frac { 1 }{ m } +\frac { 1 }{ n }

\frac { 1 }{ mn }

2 solutions

Personal Data
Apr 16, 2015

Write the ${ r }^{ th }$ term as ${ T }_{ r }=a+b\left( r-1 \right)$ . We can also assume that m>n.

Then we have that ${ T }_{ m }={ T }_{ n }+b\left( m-n \right)$ so $\frac { 1 }{ n } =\frac { 1 }{ m } +b\left( m-n \right) \Rightarrow b=\frac { 1 }{ mn }$ . We also have that ${ T }_{ m }=a+b\left( m-1 \right)$ so $\frac { 1 }{ n } =a+\frac { 1 }{ mn } (m-1)\Rightarrow a=\frac { 1 }{ mn }$ . Finally we conclude that ${ T }_{ r }=\frac { 1 }{ mn } +\frac { 1 }{ mn } (r-1)=\frac { 1 }{ mn } r$ therefore ${ T }_{ mn }=\frac { 1 }{ mn } \cdot mn=1$ .

Harvindersingh Chavan
Apr 21, 2015

Let the series be 1/2, 1, 3/2 ........ Now 1st term is 1/2 and 2nd is 1/1. So m is 1 and n is 2. Mnth term is 2nd term i.e. 1

Arithmetically correct

Practice for Bitsat here .

2 solutions

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