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1. I need to rearrange this equation to find 'r' but I'm finding it a bit hard...

8 = 3/1-r

Also, there is a question which asks "A geometric series has first term 54 an 4th term 2. What is its common ratio?"

Thanks.
2. kwalia.

So what do you get if we multiply both sides of the equation by 1-r?

a geometric series has first term 54 and 4th term 2.

So we have that a=54 and that ar^3=2 can you solve this?
3. I dont know, thats why im stuck...
4. 8=3/(1-r), with r different of 1

Multiply each side by (1-r)
you get 8(1-r) = 3(1-r) / (1-r)

On the RHS, the (1-r) cancels out

Develop the LHS, and you get 8-8r=3, so 8r=5 so r=5/8
5. And for your second question, your notations are quite confusing, but basically, if
a=54
a*r^3 = 2,

then you have r^3=2/a = 1/27
so r= cube root (1/27)

You know that 3^3 = 27 (or you find it)
Therefore, r=1/3

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Updated: February 7, 2010
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