Question

If the $m^{th}$ term of an A.P be $\frac{1}{n}$ and $n^{th}$ term be $\frac{1}{m}$, then find the $(10mn{)}^{th}$ term of an A.P.

• A. 9
• B. 8
• C. 10
• D. 11

Answer 1

The correct option is C 10

Let the first term be 'a' and common difference be 'd'
$t_m=\frac{1}{n}$
$a+(m-1)d=\frac{1}{n}$ ...............(1)
$t_n=\frac{1}{m}$
$a+(n-1)d=\frac{1}{m}$ ...............(2)

Subtracting (2) from (1), we get:
$a+(m-1)d=\frac{1}{n}$
$a+(n-1)d=\frac{1}{m}$
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$\left[\right(m-1)-(n-1\left)\right]d=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow (m-1-n+1)d=\frac{m-n}{mn}$
$\Rightarrow (m-n)d=\frac{(m-n)}{mn}$
$d=\frac{1}{mn}$

Substituting $d=\frac{1}{mn}$ in (1), we get:
$a+(m-1)d=\frac{1}{n}$
$\Rightarrow a+(m-1)\frac{1}{mn}=\frac{1}{n}$
$\Rightarrow a=\frac{1}{n}-\frac{(m-1)}{mn}=\frac{m-m+1}{mn}$
$a=\frac{1}{mn}$
$t_{10mn}=a+(10mn-1)d$
$=\frac{1}{mn}+\frac{(10mn-1)}{mn}$
$=\frac{1+10mn-1}{mn}=\frac{10mn}{mn}$
$t_{10mn}=10$
Thus, $(10mn{)}^{th}$ term is 10