8. Sequences and Series
easy

Let ${T_r}$ be the ${r^{th}}$ term of an $A.P.$ 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

A

$\frac{1}{{mn}}$

B

$\frac{1}{m} + \frac{1}{n}$

C

$1$

D

$0$

(IIT-1998)

Solution

(c) ${T_m} = a + (m – 1)\,d = \frac{1}{n}$

and ${T_n} = a + (n – 1)\,d = \frac{1}{m}$

On solving $a = \frac{1}{{mn}}$ and $d = \frac{1}{{mn}}$

$\therefore $ ${T_{mn}} = a + (mn – 1)\,d = \frac{1}{{mn}} + (mn – 1)\frac{1}{{mn}} = 1$

Standard 11
Mathematics

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