Question 11 Exercise 4.4

Solutions of Question 11 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that the prodect of $\mathrm{n}$ geometric means between $a$ and $b$ is equal to the $nth$ power for the single geometric mean between them.

Let $G_1, G_2, G_9, \ldots, G_n$ be the $n$ geometric means between $a$ and $b$,
then $a, G_1, G_2, G_3, \ldots, G_n, b$ is a geometric sequence having $n+2$ terms, where $a_{n+2}=b$.
We know that $a_n=a_1 r^{n-1}$, replacing $n$ by $n+2$
$$\begin{align}a_{n+2}&=a_1 r^{n i 1}=a r^{n+1}=b \\ \because a_1&=a \\ \Rightarrow \quad r^{n+1}&=\dfrac{b}{a} . \\ \Rightarrow \quad r&=(\dfrac{b}{a})^{\dfrac{1}{n+1}}\end{align}$$ Now $$\begin{align}G_1&=a_2=a r=a(\dfrac{b}{a})^{\dfrac{1}{n+1}}\\ G_2&=a_3=a_1 r^2=a(\dfrac{b}{a})^{\dfrac{2}{n+1}} \\ G_3&=a_4=a_1 r^3=a(\dfrac{b}{a})^{\dfrac{3}{n+1}}\\ .\\ .\\ .\\ G_n&=a_{n+1}=a_1 r^n=a(\dfrac{b}{a})^{\dfrac{n}{n+1}}\\ G_n&=a_{n+1}=a_1 r^n=a(\dfrac{b}{a})^{\dfrac{n}{n+1}} . \end{align}$$ $$\begin{align}G_1 \cdot G_2 \cdot G_3 \ldots G_n&=a(\dfrac{b}{a})^{\dfrac{1}{n+1}} \cdot a(\dfrac{b}{a})^{\dfrac{2}{n+1}} \cdot a(\dfrac{b}{a})^{\dfrac{3}{n+1}}\ldots a(\dfrac{b}{a})^{\dfrac{n}{n+1}}\\ & =a^n \cdot(\dfrac{b}{a})^{\dfrac{1}{n+1}+\dfrac{2}{n+1}+\dfrac{3}{n+1}+\ldots+\dfrac{n}{n+1}} \\ & =a^n(\dfrac{b}{a})^{\dfrac {1}{n+1}(1+2+3+\ldots+n)} \\ & =a^n(\dfrac{b}{a})^{\dfrac{1}{n+1} \cdot \dfrac{n(n+1)}{2}} \quad\because (1+2+3+\ldots+n)=\dfrac{n(n+1)}{2} \\ & =a^n(\dfrac{b}{a})^{\dfrac{n}{2}}\\ &=a^n(\dfrac{b^{\dfrac{n}{2}}}{a^\dfrac{n}{2}}) \\ & =a^{n-\dfrac{n}{2}} b^{\dfrac{n}{2}}\\ &=a ^\dfrac{2 n-n}{2} b^{\dfrac{n}{2}} \\ & =a^{\dfrac{n}{2}} b^{\dfrac{n}{2}}\\ &=(a b)^{\dfrac{n}{2}} \\ \Rightarrow G_1 \cdot G_2 \cdot G_3 \ldots G_n&=((a b)^{\dfrac{1}{2}})^n \\ \Rightarrow G_1 \cdot G_2 \cdot G_3 \ldots G_n&=(\sqrt{a b})^n=G^n \\ \because G&=\sqrt{a b}.\end{align}$$ Which is the required result.

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