progression problem

[Valentas]

I've got a problem;

Three different numbers create geometric progression. When we multiply them in order by 1, 2 and 3 we get a arithmetic progression. I need to find out geometric progression denominator.

Well I understand like this:

b1 ; b2 ; b3 - geometric progression

b1; 2b2 ; 3b3 - arithmetic progression

But I can't quite understand how to find out the denominator because I get it q = 0 ; Am I right or I missed something?

 

[Deleted member 4993]

I've got a problem;

Three different numbers create geometric progression. When we multiply them in order by 1, 2 and 3 we get a arithmetic progression. I need to find out geometric progression denominator.

Well I understand like this:

b1 ; b2 ; b3 - geometric progression

b1; 2b2 ; 3b3 - arithmetic progression

But I can't quite understand how to find out the denominator because I get it q = 0 ; Am I right or I missed something?

I am getting different answer.

I assume by "progression denominator" - it was meant "common ratio"(=r as in a, ar, ar[sup:32o89mfi]2[/sup:32o89mfi], ar[sup:32o89mfi]3[/sup:32o89mfi], ....).

Please share your work - or explain the term "progression denominator".

 

[TchrWill]

I've got a problem;

Three different numbers create geometric progression. When we multiply them in order by 1, 2 and 3 we get a arithmetic progression. I need to find out geometric progression denominator.

Well I understand like this:

b1 ; b2 ; b3 - geometric progression

b1; 2b2 ; 3b3 - arithmetic progression

But I can't quite understand how to find out the denominator because I get it q = 0 ; Am I right or I missed something?

If I read you correctly:

Let the geometric progression be
a, ar, ar^
Then, the arithmetic progression becomes
a, 2ar, 3ar^2

Subtracting successive terms, we get
2ar - a = 3ar^2 - 2ar
The "a's" cancel out leaving us with 3r^2 - 4r + 1 = 0

Then, using the quadratic equation, r = [4+/-sqrt(16 - 12)]/2 = [4+/-2]/12
Therefore, r = [4 + 2]/6 = 6/6 = 1 or [4 - 2]/6 = 2/6 = 1/3.

 

[soroban]

Hello, Valentas!

Three different numbers create a geometric progression.
When we multiply them in order by 1, 2 and 3 we get an arithmetic progression.
I need to find out geometric progression denominator.

\(\displaystyle \text{The three terms of the geometric sequence are: }\:a,\;ar,\;ar^2\)
. . \(\displaystyle \text{Note: since these are }di\!f\!f\!erent\text{ numbers, }\,r \ne 1.\)

\(\displaystyle \text{The three terms of the arithmetic sequence are: }\:a,\;2ar,\;3ar^2\)

\(\displaystyle \text{Let }d\text{ be their common difference.}\)

\(\displaystyle \text{Then we have: }\:\begin{Bmatrix}a+d &=& 2ar & \Rightarrow & d &=& 2ar - a & [1] \\ 2ar + d &=& 3ar^2 & \Rightarrow & d &=& 3ar^2 - 2ar & [2] \end{Bmatrix}\)

\(\displaystyle \text{Equate [2] and [1]: }\;3ar^2-2ar \:=\:2ar-a \quad\Rightarrow\quad 3r^2 - 4r + 1\:=\:0\)

. . \(\displaystyle (r - 1)(3r - 1) \:=\:0 \quad\Rightarrow\quad \rlap{/////}r\,=\,1,\;\;r \,=\,\tfrac{1}{3}\)

 

[Valentas]

Thank you guys a lot