Let $T_{r}$, be $r$th term of an AP whose first term is $a$ and common difference is $d$. If for some positive integers $m,n,m
eq n,\ T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m}$, then $a - d$ equals

Question

Let $T_{r}$, be $r$th term of an AP whose first term is $a$ and common difference is $d$. If for some positive integers $m,n,m 
eq n,\ T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m}$, then $a - d$ equals

Dr. Abubacker · Accepted Answer

(a)
Since, $T_{m} = \frac{1}{n} \Rightarrow a + (m - 1)d$ …(i)
and $T_{n} = \frac{1}{m} = a + (n - 1)d$ …(ii)
On solving Eqs. (i) and (ii), we get
$a = \frac{1}{mn}$ and $d = \frac{1}{mn}$
$	herefore\ a - d = \frac{1}{mn} - \frac{1}{mn} = 0$

Q. Let $T_{r}$, be $r$th term of an AP 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

Explanation

Reviewed by Dr. Abubacker

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