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If $n$ geometric means be inserted between $a$ and $b$ then the $n^{\text {th }}$ geometric mean will be
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Verified Answer
The correct answer is:
$a\left(\frac{b}{a}\right)^{\frac{n}{n+1}}$
If $n$ geometric means $g_1, g_2 \ldots \ldots g_n$ are to be inserted between two positive real numbers $a$ and $b$, then $a, g_1, g_2 \ldots \ldots . g_n, b$ are in G.P. Then
$g_1=a r, g_2=a r^2 \ldots \ldots . . . g_n=a r^n$
So, $b=a r^{n+1} \Rightarrow r=\left(\frac{b}{a}\right)^{1 /(n+1)}$
Now $n^{\text {th }}$ geometric mean $\left(g_n\right)=a r^n=a\left(\frac{b}{a}\right)^{n /(n+1)}$
Aliter : As we have the $m^{\text {th }}$ G.M. is given by
$G_m=a\left(\frac{b}{a}\right)^{\frac{m}{n+1}}$
Now replace $m$ by $n$ we get the required result.
$g_1=a r, g_2=a r^2 \ldots \ldots . . . g_n=a r^n$
So, $b=a r^{n+1} \Rightarrow r=\left(\frac{b}{a}\right)^{1 /(n+1)}$
Now $n^{\text {th }}$ geometric mean $\left(g_n\right)=a r^n=a\left(\frac{b}{a}\right)^{n /(n+1)}$
Aliter : As we have the $m^{\text {th }}$ G.M. is given by
$G_m=a\left(\frac{b}{a}\right)^{\frac{m}{n+1}}$
Now replace $m$ by $n$ we get the required result.
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