Question

Let $T_r$ be the $r^{th}$ term of an A.P. for $r=1,2,3,\dots$ If for some positive integers $m,n$, $T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $T_{mn}$ equals

• A. $\frac{1}{mn}$
• B. $\frac{1}{m}+\frac{1}{n}$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Let $a$ be the first term and $r$ be the common difference of A.P.
$T_m=a+(m-1)d=\frac{1}{n}\cdots \left(1\right)T_n=a+(n-1)d=\frac{1}{m}\cdots \left(2\right)$

Subtracting $\left(1\right)$ and $\left(2\right)$,
$(m-n)d=\frac{1}{n}-\frac{1}{m}=\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn}$
Putting $d=\frac{1}{mn}$ in eqn$\left(1\right)$, we get $a=\frac{1}{mn}$

$\therefore T_{mn}=a+(mn-1)d\Rightarrow T_{mn}=\frac{1}{mn}+(mn-1)\frac{1}{mn}\Rightarrow T_{mn}=\frac{1}{mn}+1-\frac{1}{mn}\Rightarrow T_{mn}=1$