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8. Sequences and Series
easy
If the geometric mean between $a$ and $b$ is $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$, then the value of $n$ is
A
$1$
B
$-1/2$
C
$1/2$
D
$2$
Solution
(b) As given $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = {(ab)^{1/2}}$
$ \Rightarrow $ ${a^{n + 1}} – {a^{n + 1/2}}{b^{1/2}} + {b^{n + 1}} – {a^{1/2}}{b^{n + 1/2}} = 0$
$ \Rightarrow $ $({a^{n + 1/2}} – {b^{n + 1/2}})({a^{1/2}} – {b^{1/2}}) = 0$
$ \Rightarrow $ ${a^{n + 1/2}} – {b^{n + 1/2}} = 0$
$(\because \;a \ne b \Rightarrow {a^{1/2}} \ne {b^{1/2}})$
$ \Rightarrow $ ${\left( {\frac{a}{b}} \right)^{n + 1/2}} = 1 = {\left( {\frac{a}{b}} \right)^0} $
$\Rightarrow n + \frac{1}{2} = 0$
$\Rightarrow n = – \frac{1}{2}$.
Standard 11
Mathematics
