Question

Let $T_r$ be the $r$th term of an AP whose first term is $a$ and common difference is $d$. If for some positive integers $m,n,m\ne n,T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $a-d$ is equal to

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

Answer 1

The correct option is A $0$

Given, $T_m=\frac{1}{n}\Rightarrow a+(m-1)d=\frac{1}{n}$ ....(i)
and $T_n=\frac{1}{m}\Rightarrow a+(n-1)d=\frac{1}{m}$ ...(ii)
subtract (i) and (ii), we get

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

substitute $d$ in (i)

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

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

Now, $a-d=\frac{1}{mn}-\frac{1}{mn}=0$