Question

How to find four geometric means between 243 and 1?

Answer 1

the geometric mean between a and b is sqrt(ab)

Answer 2

sqrt(ab)?

Answer 3

You can treat 243 as \(a_1\) of a geometric series and 1 as \(a_5\) and use the formula for the \(n^{th}\) term of a geometric series to find r.

Answer 4

I dont know the equation for geometric series..

Answer 5

\(a_n = a_1 * r^{n-1}\)

Answer 6

\(a_5 = 1 = a_1 * r^{5-1} = 243 * r^4\)

Answer 7

So r is equal to the fourth root of \(\frac{1}{243}\)

Answer 8

To get 4 means in between you will need to set 1 = a_6 instead though, sorry.

Answer 9

Which will mean the fifth root instead of the fourth root.

Answer 10

a5= 1 and a1= 1?

Answer 11

\[a_n = 243 * \left(\sqrt[5]{\frac{1}{243}}\right)^{n-1}\]

Answer 12

It should be \(a_1\) = 243 and \(a_6=1\).

Answer 13

how do we know it is the 6th term?

Answer 14

243 is the first term, and you want 4 terms to between 243 and one. So the last term you're trying to find will be the fifth one, and 1 is then the 6th term.

Answer 15

Like this: 243, \(a_2\), \(a_3\), \(a_4\), \(a_5\), 1

Answer 16

so i do \[a _{5}=243(r ^{5-1})\] right?

Answer 17

You could do that to find the fifth term, but because you know the sixth term in order to find r you'll want to use 6 instead of 5.

Answer 18

oh! yeah..

Answer 19

Because you know \(a_6 = 1\) you can solve for r if you set \(n=6\).

Answer 20

so why is it like you said earlier? the \[\frac{ 1 }{ 243 }\]

Answer 21

i know it is \[a _{6}=243 \times (r ^{5})\] but i dont know how to solve it completely..

Answer 22

When you set \(n=6\), you get \(a_6 = a_1 * r^{n-1}\), then \( 1 = 243 * r^5 \).

Answer 23

where did the 1 come from?

Answer 24

oh nevermind!

Answer 25

You know \(a_6 = 1\).

Answer 26

okay so now how do i finish solving it?

Answer 27

wait i get it.. but what does it come out to if i dont have a calculator?

Answer 28

\[r=\sqrt[5]{\frac{ 1 }{ 243 }}\] is what i got it down to

Answer 29

\(\sqrt[5]{243} = 3\) so you can simplify it if you split the fraction up.

Answer 30

how did you split it up? ive never done that

Answer 31

and where did r go?

Answer 32

That was just a general statement, not part of the equation.

Answer 33

I was just saying that you can simplify that root down to a nice number.

Answer 34

yes but how did you do that

Answer 35

You could then say that \(r = \frac{1}{3}\).

Answer 36

\(\sqrt{ \frac{a}{b} } = \frac{\sqrt{a}}{\sqrt{b}}\) when a and b are both positive.

Answer 37

ok but if i kept just what i had without simplifiying it down how would i do it?

Answer 38

The easiest way would just be to leave everything in root form in the answer.

Answer 39

ok!

Answer 40

i.e. \(a_2 = 243 * \sqrt[5]{\frac{1}{243}}\)

Answer 41

How much you're expected to simplify would really depend on the class and what unit you're doing etc. so it's hard to say exactly how far your particular teacher will want it simplified.

Answer 42

do you know the formula for the sum of geometric series? i cant find it

Answer 43

\[a_1 \frac{1-r^n}{1-r}\]

Answer 44

thank you