# Math Help - Geometric Progression

1. ## Geometric Progression

Hello I have the follow question from a practise revision test that I am strugling with,

The main problem I am having is finding the value of r in my equation...

Here is the question...

A lathe has a range of six speeds. the lowest beign 81 RPM and the highest beign 954 RPM. Determine the six speeds if they are arranged in geometric progression.

Thanks in advance for any help.

Kind regards,
Chris

2. Originally Posted by HNCMATHS
Hello I have the follow question from a practise revision test that I am strugling with,

The main problem I am having is finding the value of r in my equation...

Here is the question...

A lathe has a range of six speeds. the lowest beign 81 RPM and the highest beign 954 RPM. Determine the six speeds if they are arranged in geometric progression.

Thanks in advance for any help.

Kind regards,
Chris
the six speeds ...

$81, 81r, 81r^2, 81r^3, 81r^4, 81r^5$

so ...

$81r^5 = 954$

solve for $r$ ... it will not be a "nice" solution.

3. Originally Posted by HNCMATHS
Hello I have the follow question from a practise revision test that I am strugling with,

The main problem I am having is finding the value of r in my equation...

Here is the question...

A lathe has a range of six speeds. the lowest beign 81 RPM and the highest beign 954 RPM. Determine the six speeds if they are arranged in geometric progression.

Thanks in advance for any help.

Kind regards,
Chris
Okay, do you know what a "geometric progression" is? It is a sequence of numbers, each of which (after the first number) is a constant multiple of the previous number.

If the first number is "a" and the constant multiple is "r" then the sequence is a, ar, $ar^2$, $ar^3$, $ar^4$, $ar^5$, etc.

Since you know the lowest speed is 81 and the highest is 954, you know that a= 81 and $ar^5= 954$. Solve for a and r, then calculate the four speeds ar, $ar^2$, $ar^3$, and $ar^4$.

(Arrrgh, Skeeter got in ahead of me again. Blast you!)

4. Im my notes it says that...

$
r= (n-1) sqrt b/a
$

5. Originally Posted by HallsofIvy
Okay, do you know what a "geometric progression" is? It is a sequence of numbers, each of which (after the first number) is a constant multiple of the previous number.

If the first number is "a" and the constant multiple is "r" then the sequence is a, ar, $ar^2$, $ar^3$, $ar^4$, $ar^5$, etc.

Since you know the lowest speed is 81 and the highest is 954, you know that a= 81 and $ar^5= 954$. Solve for a and r, then calculate the four speeds ar, $ar^2$, $ar^3$, and $ar^4$.

(Arrrgh, Skeeter got in ahead of me again. Blast you!)
I still dont understand how we get the value for r

6. anyone ?

7. Originally Posted by HNCMATHS
I still dont understand how we get the value for r
be patient ... don't bump.

$81r^5 = 954$

$r^5 = \frac{954}{81} = \frac{106}{9}$

$r = \left(\frac{106}{9}\right)^{\frac{1}{5}} = \sqrt[5]{\frac{106}{9}}$

8. Originally Posted by skeeter
be patient ... don't bump.

$81r^5 = 954$

$r = \left(\frac{106}{9}\right)^{\frac{1}{5}} = \sqrt[5]{\frac{106}{9}}$

Hello Skeeter, Sorry about that ...Why does

$r^5 = \frac{954}{81} = \frac{106}{9}$

9. Unfortunately I am now totally lost Sorry

10. so r = 17.159 ?

11. Originally Posted by HNCMATHS
Im my notes it says that...

$
r= (n-1) sqrt b/a
$
What was that referring to? Since your original problem had NO a, b, or n, I don't see how it applies.

Seriously, this is just basic algebra- except for the definition of "geometric sequence" and fifth root, really just arithmetic.

Skeeter and I told you that if the first term in a geometric series is 'a' and "constant ratio" is 'r' then the sixth term is $ar^5$. Here you are told that the first term is 81, so a= 81, and the six term is 954 so $81r^5= 954$. Then, as skeeter said, $r^5= \frac{954}{81}= \frac{106}{9}$, by cancelling a "9" in both numerator and denominator. To get rid of that 5th power, take the fifth root: $r= \sqrt[5]{\frac{106}{9}= 11.777777... = 1.638$ approximately. Not anything near "17.159"- that's larger than 106/9 already! The fifth root of a number larger than 1 will be much less that the number.

Now that you know that r= 1.638, the four values you need to find are (81)(1.638), $(81)(1.638)^2$, $(81)(1.638)^3$, and $(81)(1.638)^4$.

12. Originally Posted by HallsofIvy
What was that referring to? Since your original problem had NO a, b, or n, I don't see how it applies.

Seriously, this is just basic algebra- except for the definition of "geometric sequence" and fifth root, really just arithmetic.

Skeeter and I told you that if the first term in a geometric series is 'a' and "constant ratio" is 'r' then the sixth term is $ar^5$. Here you are told that the first term is 81, so a= 81, and the six term is 954 so $81r^5= 954$. Then, as skeeter said, $r^5= \frac{954}{81}= \frac{106}{9}$, by cancelling a "9" in both numerator and denominator. To get rid of that 5th power, take the fifth root: $r= \sqrt[5]{\frac{106}{9}= 11.777777... = 1.638$ approximately. Not anything near "17.159"- that's larger than 106/9 already! The fifth root of a number larger than 1 will be much less that the number.

Now that you know that r= 1.638, the four values you need to find are (81)(1.638), $(81)(1.638)^2$, $(81)(1.638)^3$, and $(81)(1.638)^4$.
Thank you very much for your help! Now i have worked out how to take the fifth root using my calculator this all makes sense,

Thank you very much,

Best Regards,
Chris