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8. Sequences and Series
easy
${7^{th}}$ term of the sequence $\sqrt 2 ,\;\sqrt {10} ,\;5\sqrt 2 ,\;.......$ is
A
$125\sqrt {10} $
B
$25\sqrt 2 $
C
$125$
D
$125\sqrt 2 $
Solution
(d) Given sequence is $\sqrt 2 ,\;\sqrt {10} ,\;\sqrt {50} ……..$
Common ratio $r = \sqrt 5 $,
first term $a = \sqrt 2 $, then ${7^{th}}$ term
${t_7} = \sqrt 2 {(\sqrt 5 )^{7 – 1}} = \sqrt 2 {(\sqrt 5 )^6}$
$= \sqrt 2 {(5)^3} = 125\sqrt 2 $.
Standard 11
Mathematics
