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

G_{1} . G_{2} . . . . . . . G_{n} = G

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G_{1} . G_{2} . . . . . . . G_{n} = G^{1 / n}

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G_{1} . G_{2} . . . . . . . G_{n} = G^{n}

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G_{1} . G_{2} . . . . . . . G_{n} = G^{2 / n}

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Solution

The correct option is C $G_{1} . G_{2} . . . . . . . G_{n} = G^{n}$
Here $G = (a b)^{1 / 2} a n d G_{1} = a r^{1}, G_{2} = a r^{2}, . . . . . . G_{n} = a r^{n}$
Therefore $G_{1} . G_{2} . G_{3} . . . . . . G_{n} = a^{n} r^{1 + 2 + . . . . + n} = a^{n} r^{n (n + 1) / 2}$
But $a r^{n + 1} = b \Rightarrow r = {(\frac{b}{a})}^{1 / (n + 1)}$
Therefore, the required product is
$a^{n} {(\frac{b}{a})}^{1 / (n + 1) . n (n + 1) / 2} = (a b)^{n / 2} = {(a b)^{1 / 2}}^{n} = G^{n} .$

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G_{1}, G_{2}, . . . . . . . G_{n}

If n geometric means between a and b be $G_{1}, G_{2}, . . . . G_{n}$ and a geometric mean of a and b be G, then the true relation is

G_{1}, G_{2}, . . . . . . . G_{n}

If G₁, G₂, ..., G_n are n geometric means between a and b, then common ratio is:

G_{1}, G_{2}

\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}

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