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Let be the $r^{th}$ term of an A.P. whose first term is a and common difference is d.

If for some positive integers $m,n,m\neq n,T_{m}=\frac{1}{n},\: and\: T_{n}=\frac{1}{m},$ then $a-d$ equals
Option: 1
$1/mn$

Option: 2
$1$

Option: 3
$0$

Option: 4
$\frac{1}{m}+\frac{1}{n}$

Answers (1)

best_answer

Use the concept of

General term of an A.P.

$T_{n}= a+\left ( n-1 \right )d$

$a\rightarrow$ First term

$n\rightarrow$ number of term

$d\rightarrow$ common difference

Given:

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

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

Subtracing above 2 equations

$\therefore \:\frac{1}{n}-\frac{1}{m}=(m-n)d$

$\therefore \:\frac{m-n}{mn}=(m-n)d$

$\therefore \: d=\frac{1}{mn}$

From (i) $\frac{1}{n}=a+\frac{m-1}{mn}$

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

$\therefore a-d=\frac{1}{mn}-\frac{1}{mn}=zero$

Posted by

Divya Prakash Singh

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