If there are n geometric means between a and | vedclass

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(A) Let the $n$ geometric means be $G_1, G_2, \dots, G_n$. Then the sequence $a, G_1, G_2, \dots, G_n, b$ forms a geometric progression with $n+2$ ter

Vedclass · Accepted Answer

$(\frac{b}{a})^{\frac{1}{n+1}}$

Vedclass · Answer

(A) Let the $n$ geometric means be $G_1, G_2, \dots, G_n$. Then the sequence $a, G_1, G_2, \dots, G_n, b$ forms a geometric progression with $n+2$ terms. Let $r$ be the common ratio. The $(n+2)$-th term is given by $T_{n+2} = a \cdot r^{(n+2)-1} = a \cdot r^{n+1}$. Since $T_{n+2} = b$,we have $b = a \cdot r^{n+1}$. Therefore,$r^{n+1} = \frac{b}{a}$. Thus,$r = (\frac{b}{a})^{\frac{1}{n+1}}$.

If there are $n$ geometric means between $a$ and $b$,what is the common ratio?

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