Need Help on Geometric Sequence

[hi25125]

you were given the numbers 1 to 20, and told to form geometric sequences from these given numbers.
(for examples: 1,3,9 or 1,4,16 or 2,4,8 )
What is the maximum common ratio you can use for these sequences?
Please explain your reasoning. Hint: this question can be answered beautifully by using the geometric mean.


firstable, I am a little confused about the question being asked, not really sure what it means.
I have no idea to how to begin solving this question but I know some facts that might be related to this question.
geometri mean's formula is B^2=AC
Maximum Common Ratio might have something to do with the this formula.
I am stuck here and don't know how to continue. Someone please help me. Appreciated very much.

 

[tkhunny]

Geometric sequence

a = first value

a*r = second value

a^r^r = third value

etc.

\(\displaystyle \frac{ThirdValue}{SecondValue} = \frac{a\cdot r^2}{a\cdot r} = r\)

\(\displaystyle \frac{SecondValue}{FirstValue} = \frac{a\cdot r}{a} = r\)

I'm seeing a pattern.

 

[soroban]

Hello, hi25125!

The wording of the problem is downright sloppy!

You are given the numbers 1 to 20, and told to form geometric sequences from these given numbers.
(For examples: 1,3,9 or 1,4,16 or 2,4,8 )
What is the maximum common ratio you can use for these sequences?
Please explain your reasoning.

Hint: this question can be answered beautifully by using the geometric mean.
. . I don't understand this.

First, I am a little confused about the question being asked, not really sure what it means. .Me, too!
I have no idea to how to begin solving this question but I know some facts that might be related to this question.
Geometric mean's formula is: \(\displaystyle B^2=AC\)

Maximum Common Ratio might have something to do with the this formula.
I am stuck here and don't know how to continue.

\(\displaystyle \text{If }A, B, C\text{ are three consecutive terms of a geomtric sequence,}\)
. . \(\displaystyle \text{it is always true that: }\,B^2 = AC\)
\(\displaystyle \text{But I don't see how that helps us find the common ratio.}\)
. . \(\displaystyle \text{It's easier to divide }B\text{ by }A.\)

\(\displaystyle \text{Using the integers from 1 to 20, we can form these geometric sequences:}\)

. . \(\displaystyle \begin{array}{cc}1, 2, 4, 8, 16, \rlap{//}32 & r=2\\ \\[-4mm] 1, 3, 9, \rlap{//}27, \rlap{//}81 & r=3 \\ \\[-4mm] 1, 4, 16, \rlap{//}64, \rlap{///}256 & r=4 \\ \\[-4mm] 1, 5, \rlap{//}25, \rlap{///}125 & r=5 \\ \\[-4mm] 1, 6, \rlap{//}36, \rlap{///}216 & r=6 \\ \\[-3mm] \vdots & \vdots \\ \\[-3mm] 1,19, \rlap{///}361, \rlap{////}6859 & r=19 \\ \\[-4mm] 1,20, \rlap{///}400, \rlap{////}8000 & r=20 \end{array}\) . . . \(\displaystyle \begin{array}{cc}2,4,8,16,\rlap{//}32 & r=2 \\ \\[-4mm] 2,6,18,\rlap{//}54 & r=3 \\ \\[-4mm] 2,8, \rlap{//}32 & r=4 \\ \\[-4mm] 2,10,\rlap{//}50 & r=5 \\ \\[-4mm] 2,12,\rlap{//}72 & r=6 \\ \\[-4mm] 2,14,\rlap{//}98 & r=7 \\ \\[-4mm] 2,16,\rlap{///}128 & r=8 \\ \\[-4mm] 2,18, \rlap{///}162 & r=9 \\ \\[-4mm] 2,20\,\rlap{///}200 & r=10 \end{array}\quad\hdots\text{ etc.}\)

It seems that the maximum common ratio is: \(\displaystyle r = 20\)

Is that what they're looking for?
. . The question is kind of silly, isn't it?