8. Sequences and Series
easy

माना $r = 1,\;2,\;3,....$ के लिये एक समान्तर श्रेणी का $r$ वाँ पद ${T_r}$ है। यदि किन्हीं धनात्मक पूर्णांकों $m,\;n$ के  लिये ${T_m} = \frac{1}{n}$ और ${T_n} = \frac{1}{m}$ हों, तो ${T_{mn}}$ का मान होगा

A

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

B

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

C

$1$

D

$0$

(IIT-1998)

Solution

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

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

हल करने पर,  $a = \frac{1}{{mn}}$ एवं  $d = \frac{1}{{mn}}$

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

Standard 11
Mathematics

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