Question

If the $m^{th}$ term of an A.P be $\frac{1}{n}$ ,$n^{th}$ term be $\frac{1}{m}$ , show that it's ${mn}^{th}$ term is 1.

Answer 1

As we know that, $r^{th}$ term in an A.P. is given as-

$a_r=a+(r-1)d$

Whereas, $a$ and $d$ are the first term and common difference

Therefore,

$a_m=\frac{1}{n}\left(\text{Given}\right)$

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

$a_n=\frac{1}{m}\left(\text{Given}\right)$

$a+(n-1)d=\frac{1}{m}.....\left(2\right)$

Subtracting ${eq}^n\left(2\right)$ from $\left(1\right)$, we have

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

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

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

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

Substituting the vaue of $d$ in ${eq}^n\left(1\right)$, we have

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

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

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

Now,

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

$\Rightarrow a_{mn}=\frac{1}{mn}+(mn-1)\frac{1}{mn}$

$\Rightarrow a_{mn}=\frac{1}{mn}+1-\frac{1}{mn}=1$

Hence it is proved that the ${mn}^{th}$ term of the A.P. is $1$.