If G1 is the first of n GMs between two positive numbers a and b , then show that G1 - Maths - Sequences and Series

Question

If G1 is the first of n GMs between two positive numbers a and b
, then show that G1 to the power n+1 = a tothe power n*b

Mayank Bhardwaj · Accepted Answer

Dear Student!

Here is the answer to your query.

Given : $G_{1}$ is first of n G. M.'s between two positive number a and b.

Then the given G. P. consists of ( n + 2) terms

Let r be the common ratio of this G.P. Then,

$b = ({n + 2)}^{th} term = a r^{n +1} \Rightarrow r^{n + 1} = \frac{b}{a} \Rightarrow r = (\frac{b}{a})^{\frac{1}{n + 1}} ∴ G_{1} = a r = a (\frac{b}{a})^{\frac{1}{n + 1}} \Rightarrow (G_{1})^{n + 1} = a^{n + 1} \times \frac{b}{a} = a^{n} b$

Cheers!

If G1 is the first of n GMs between two positive numbers a and b , then show that G1 to the power n+1 = a tothe power n*b.