Question

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

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

Answer 1

Answer: (C)

1

Answer 2

Let \hspace20mm T_m = a +(m-1) d = \frac{1}{n} \hspace20mm ...(i)
and \hspace20mm T_n = a +(n-1) d = \frac{1}{m} \hspace20mm ...(ii)
on substracting E (ii) from E (i), we get
$\, \, \, (m-n) \, d= \frac{1}{n} - \frac{1}{m} = \frac{m-n}{mn}$
\Rightarrow \hspace10mm d=\frac{1}{mn}
Again, $T_{mn} = a\, +\, (mn-1)\, d = a \, + (mn-n+n-1) \, d$
\hspace25mm = a\, +\, (n-1)\, d +\, (mn-n) \, d
\hspace25mm = T_n\, +n \, (m-1)\, \frac{1}{mn} = \frac{1}{m} \, + \frac{(m-1)}{m} \, = 1