Question

How do I find A1 of a geometric sequence given A4=125 and A7=1953.125?

Answer 1

\(a_{4} = a_{1}r^{3} = 125\)

\(a_{6} = a_{1}r^{6} = 1953.125\)

Answer 2

So would I then subtract?

Answer 3

You would use the two equations in any way you deemed necessary to solve the problem.

Answer 4

raise both to their respective power, no?

Answer 5

You have two equations in two unknowns.

\(a_{1}r^{3} = 125\)

\(a_{1}r^{6} = 1953.125\)

How do you find \(a_{1}\;and\;r\)? It is algebra. You've been preparing for this moment your entire mathematics career. Breakthrough the darkness and see it! How do you find the two values?

Answer 6

Thanks

Answer 7

Why would I be saying anything else? Except, of course, that I never would use the term "plug in". :-)

Answer 8

substitute?

Answer 9

Personally, I might do this:

\(a_{1}r^{6} = a_{1}r^{3}r^{3} = 125r^{3} = 1953.125\) and you are a moment away from \(r\).

Answer 10

I like that, thanks to both of you

Answer 11

errr....
I have 125r^3=1953.125
I raise both to ^(1/3)
and somehow got 125r=12.5???

Where's the error?

Answer 12

I was really torn on this problem. We want \(a_{1}\), so it is against my inclination to solve first for \(r\). However, I did want to show you that slightly odd substitution. Keep your minds open.

I would prefer to solve directly for \(a_{1}\)

I might do it like this:

\(r^{6} = \left(r^{3}\right)^{2} = \dfrac{1953.125}{a_{1}} = \left(\dfrac{125}{a_{1}}\right)^{2} \implies a_{1} = \dfrac{125^{2}}{1953.125} = 8\)

...and I NEVER have to know what \(r\) is!!

Answer 13

If you must...

\(125r^{3}=1953.125\implies r^{3} = \dfrac{1953.125}{125} = 15.625 \implies r = 2.5\)

Answer 14

r makes sense, let me run through a1 quickly

Answer 15

awesome, thanks tk!

Answer 16

No worries. It's what we do, here.

Keep your mind open!! Don't get stuck on one way of thinking!