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8. Sequences and Series
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If $n$ geometric means between $a$ and $b$ be ${G_1},\;{G_2},\;.....$${G_n}$ and a geometric mean be $G$, then the true relation is
A
${G_1}.{G_2}........{G_n} = G$
B
${G_1}.{G_2}........{G_n} = {G^{1/n}}$
C
${G_1}.{G_2}........{G_n} = {G^n}$
D
${G_1}.{G_2}........{G_n} = {G^{2/n}}$
Solution
(c) Here $G = {(ab)^{1/2}}$ and
${G_1} = a{r^1},\;{G_2} = a{r^2},……..{G_n} = a{r^n}$
Therefore ${G_1}.\;{G_2}.\;{G_3}…..{G_n} = {a^n}{r^{1 + 2 + … + n}} = {a^n}{r^{n(n + 1)/2}}$
But $a{r^{n + 1}} = b$
$\Rightarrow r = {\left( {\frac{b}{a}} \right)^{1/(n + 1)}}$
Therefore, the required product is
${a^n}{\left( {\frac{b}{a}} \right)^{1/(n + 1)\;.\;n(n + 1)/2}} = {(ab)^{n/2}} $
$= {\left\{ {{{(ab)}^{1/2}}} \right\}^n} = {G^n}$.
Standard 11
Mathematics
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