- Home
- Standard 11
- Mathematics
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