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8. Sequences and Series
easy
$x$ के किस मान के लिए ${\log _a}x + {\log _{\sqrt a }}x + {\log _{3\sqrt a }}x + ......... + {\log _{a\sqrt a }}x = \frac{{a + 1}}{2}$ होगा
A
$x = a$
B
$x = {a^a}$
C
$x = {a^{ - 1/a}}$
D
$x = {a^{1/a}}$
Solution
(d) ${\log _a}x + 2{\log _a}x + ……. + a{\log _a}x = \frac{{a + 1}}{2}$
$ \Rightarrow $ ${\log _a}x(1 + 2 + …….. + a) = \frac{{a + 1}}{2}$
$ \Rightarrow $ ${\log _a}x.\frac{{a(a + 1)}}{2} = \frac{{a + 1}}{2}$
$ \Rightarrow $ $x=a^{1/a}$.
Standard 11
Mathematics