# If $m^{th}$ term of an $A.P.$ is $\frac{1}{n}$ and $n^{th}$ term of another $A.P.$ is $\frac{1}{m}$. Then, show that $(mn)^{th}$ term is equal to $1$.

**Given:** $m^{th}$ term of an $A.P.$ is $\frac{1}{n}$ and $n^{th}$ term of another $A.P.$ is $\frac{1}{m}$.

**To do:** To show that $(mn)^{th}$ term is equal to $1$.

**Solution:**

As given, $m^{th}$ term$a_m=\frac{1}{n}$ and $n^{th}$ term $a_n=\frac{1}{m}$

Let $a$ and $d$ be the first term and common difference of the A.P. respectively.

$\Rightarrow a_{m}=a+(m-1)d=\frac{1}{n}$ ..........$( 1)$

And, $a_n=a+(n-1)d=\frac{1}{m}$ .......$( 2)$

On subtracting equation $( 1)$ by $( 2)$ we get,

$md-d-nd+d=\frac{1}{n}-\frac{1}{m}$

$\Rightarrow d(m-n)=\frac{m-n}{mn}$

$\Rightarrow d=\frac{1}{mn}$

On putting this value again in equation $(1)$ or $(2)$ we get,

$a=\frac{1}{mn}$

Let $A$ be the $mn^{th}$ term of the $A.P.$

$a+(mn-1)d=\frac{1}{mn}+1+(-\frac{1}{mn})=1$

hence proved.

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