Question

Let $T_r$ be the rth term of an A.P. 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 $a-d$ equals:

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

Answer 1

The correct option is A $0$

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

and $T_n=\frac{1}{m}\Rightarrow a+(n-1)d=\frac{1}{m}$ ...(2)

On solving (1) and (2), we get

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

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