Let $t_{r}$ 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 $a - d =$

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Detailed Solution

\begin{matrix} a + (m - 1) d = \frac{1}{n} \\ a + (n - 1) d = \frac{1}{m} \\ (m - n) d = \frac{m - n}{m n} \end{matrix}

$\Rightarrow d = \frac{1}{m n}$

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

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

$∴ a - d = 0$

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