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8. Sequences and Series
medium
The value of ${a^{{{\log }_b}x}}$, where $a = 0.2,\;b = \sqrt 5 ,\;x = \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + .........$to $\infty $ is
A
$1$
B
$2$
C
$\frac{1}{2}$
D
$4$
Solution
(d) $x = \frac{{1/4}}{{1 – (1/2)}} = \frac{1}{2}$
$\therefore $${\left( {\frac{1}{5}} \right)^{{{\log }_{\sqrt 5 }}\left( {\frac{1}{2}} \right)}} = {\left( {\frac{1}{5}} \right)^{{{\log }_5}\left( {\frac{1}{4}} \right)}} = {5^{ – {{\log }_5}{4^{ – 1}}}} = {5^{{{\log }_5}4}} = 4$.
Standard 11
Mathematics
