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 $m$, $n$, $m$ not equal $n$, $T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $a-d$ equals

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

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Finding $a-d$:

Given $T_r$ is the $r^{th}$ term of an AP where $r=1,2,3,\dots$.

Let '$a$‘ be the first term and ’$d$' be the common difference.

Let $m$ and $n$ be any two positive integers then $m^{th}$ term can be written as,

$\begin{array}{rcl}{T}_m& =& a+\left(m-1\right)d\\ \frac{1}{n}& =& a+\left(m-1\right)d\left[\text{given}\right]\end{array}$

And $n^{th}$ term can be written as,

$\begin{array}{rcl}{T}_n& =& a+\left(n-1\right)d\\ \frac{1}{m}& =& a+\left(n-1\right)d\left[\text{given}\right]\end{array}$

Finding the value of '$d$' by subtracting the $n^{th}$ term from $m^{th}$ term.

$\begin{array}{rcl}\frac{1}{n}-\frac{1}{m}& =& a+\left(m-1\right)d-\left(a+\left(n-1\right)d\right)\\ \frac{m-n}{mn}& =& a+\left(m-1\right)d-a-\left(n-1\right)d\\ \frac{m-n}{mn}& =& md-d-nd+d\\ \frac{m-n}{mn}& =& \left(m-n\right)d\\ d& =& \frac{1}{mn}\end{array}$

Finding the value of '$a$', by substituting $d$ as$\frac{1}{mn}$ in the equation $\begin{array}{rcl}\frac{1}{n}& =& a+\left(m-1\right)d\end{array}$.

$\begin{array}{rcl}\frac{1}{n}& =& a+\left(m-1\right)\frac{1}{mn}\\ \frac{1}{n}& =& a+\frac{m}{mn}-\frac{1}{mn}\\ \frac{1}{n}& =& a+\frac{1}{n}-\frac{1}{mn}\\ a& =& \frac{1}{mn}\end{array}$

To find the value of $a-d$, substitute $a$ as $\frac{1}{mn}$ and $d$ as $\frac{1}{mn}$in $a-d$.

$\begin{array}{rcl}a-d& =& \frac{1}{mn}-\frac{1}{mn}\\ & =& 0\end{array}$

Therefore, the value of $a-d$ is $0$.

Hence, the correct answer is option (B).