Let $T_{r}$ be the $r$th term of an AP for $r = 1,\ 2,\ 3$,… If for some positive integers $m,\ n,\ we\ have\ T_{m} = \frac{1}{n}\ and\ T_{n} = \frac{1}{m},\ then\ T_{mn}$ equals

Question

Dr. Abubacker · Accepted Answer

(c)
Given that, $T_{m} = a + (m - 1)d = \frac{1}{n}$ …(i)
And $T_{n} = a + (n - 1)d = \frac{1}{m}\$ …(ii)
Where $a$ and $b$ are the first term and common difference respectively.
On solving Eqs. (i) and (ii), we get
$a = \frac{1}{mn}\ and\ d = \frac{1}{mn}$
$	herefore\ T_{mn} = a + (mn - 1)d$
$= \frac{1}{m} + (mn - 1)\frac{1}{mn} = 1$

Q. Let $T_{r}$ be the $r$th term of an AP for $r = 1,\ 2,\ 3$,… If for some positive integers $m,\ n,\ we\ have\ T_{m} = \frac{1}{n}\ and\ T_{n} = \frac{1}{m},\ then\ T_{mn}$ equals

Explanation

Reviewed by Dr. Abubacker

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