Question

Let $T_{r}$ be rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m≠n, $T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m}$, then a – d equals

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

Answer 1

Answer: (D)

0

Answer 2

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

and $T_{n} = \frac{1}{m}$ ⇒ $a + (n - 1)d = \frac{1}{m}$ …..(ii)

Subtract (ii) from (i), we get $(m - n)d = \frac{1}{n} - \frac{1}{m}$

⇒ $(m - n)d = \frac{(m - n)}{mn}$ ⇒ $d = \frac{1}{mn}$, as m – n ≠ 0

$a = \frac{1}{m} - (n - 1)d = \frac{1}{m} - \frac{n - 1}{mn} = \frac{1}{mn} = d$. Therefore a – d = 0