Question

If the $m^{th}$ term of an $A.P$. is $\frac{1}{n}$ and $n^{th}$ term is $\frac{1}{m}$, then the sum of first $mn$ terms is ______.

• A. $mn+1$
• B. $\frac{mn+1}{2}$
• C. $\frac{mn-1}{2}$
• D. $\frac{mn-1}{3}$

Answer 1

Answer: (B)

$\frac{mn+1}{2}$

Given that,

$m^{th}$ term $=\frac{1}{n}$

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

$\Rightarrow an+mnd-nd=1....\left(1\right)$

$n^{th}$ term $=\frac{1}{m}$

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

$\Rightarrow am+mnd-md=1....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get

$an+mnd-nd=am+mnd-md$

$\Rightarrow a(n-m)=d(n-m)$

$\therefore a=d$

Putting value of $a=d$, we get

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

Sum of first $mn$ terms

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

$=amn+\frac{(mn{)}^{2}d}{2}-\frac{mnd}{2}$

$=\frac{mn}{2}+\frac{1}{2}$

$=\frac{mn+1}{2}$