Question

If the $m^{th}$ term and the nth term of an $AP$ are respectively $\frac{1}{n}$ and $\frac{1}{m}$, then the $(mn )^{th}$ term of the $AP$ is

• A. $\frac{1}{mn}$
• B. $\frac{m}{n}$
• C. $1$
• D. $\frac{n}{m}$

Answer 1

Answer: (C)

Let $a$ and $d$ be the first term and common difference of an $AP$.

Since, $T_m = \frac{1}{n}$

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

and $T_{n} = \frac{1}{m}$

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

On solving Eqs. $\left(i\right)$ and $\left(ii\right)$, we get

$a=-\frac{1}{mn}$ and $d= \frac{1}{mn}$

$\therefore T_{mn}= a +\left(mn-1\right)d $

$= \frac{1}{mn} +\frac{\left(mn-1\right)}{mn} = \frac{mn}{mn} = 1$