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)

$1$

Answer 2

Let us consider in A.P series

First term =a and common difference =d

mth term= $\frac1n$

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

$n^{th}$term = $\frac{1}{m}....(2)$

Now subtract (2) from (1)

$\therefore (m-1)d-(n-1) =\frac1n-\frac1m$

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

$d= \frac{1}{mn}.....(3)$

From (3) & (2) we get:

$a+(m-n)\frac {1}{mn} = \frac1n$

a=$\frac1n-\frac{m-1}{mn}$

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

sum of first ‘K’ terms = $\frac k2 [2a+(k-1) d]$

sum = $\frac{mn}{2}[2a+(mn-1)d]$

=$\frac{mn}{2}[2a+(mn-1) \frac{1}{mn}]=\frac{1+mn}{2}$