Kamila Frye

2022-10-18

Find 3 geometric means between 3 and 1488

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Audrey Russell

Expert

We are looking for 3 numbers a,b,c such that:
a is the geometric mean of 3 and b
b is the geometric mean of a and c
c is the geometric mean of b and 1488
That will make the following sequence into a geometric one:
3,a,b,c,1488
If the common ratio is r then we must have:
$1488=3{r}^{4}$
So:
${r}^{4}=\frac{1488}{3}=496={2}^{4}\cdot 31$
in order that the geometric means be Real and positive, we need to choose the principal 4th root to find:
$r=2\sqrt[4]{31}$
Hence a,b,c are:
$3\cdot 2\sqrt[4]{31}=6\sqrt[4]{31}$
$6\sqrt[4]{31}\cdot 2\sqrt[4]{31}=12\sqrt{31}$
$12\sqrt{31}\cdot 2\sqrt[4]{31}=24{\left(\sqrt[4]{31}\right)}^{3}$

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