Geometric Sequence: 1st 3 terms sum to 4.88, and the sum to infinity is 10

[Banks]

Hi all

If you're told that the sum of the first three terms in a geometric sequence equates to 4.88, and the sum to infinity is 10, then how would we find 'a' & 'r'?

I know the two equations needed are:

• a + ar + ar^2 = 4.88
• a / 1 - r = 10

So I thought that initially I would substitute the second equation with 'a' as the subject ( a = 10(1-r) ) into the first equation. I tried this and it was obviously wrong (I was left with -10r^3 + 10 = 4.88).

Could someone please tell me where I've gone wrong and how to solve for a and r?

Thanks in advance.

 

[lookagain]

Hi all

If you're told that the sum of the first three terms in a geometric sequence equates to 4.88, and the sum to infinity is 10, then how would we find 'a' & 'r'?

I know the two equations needed are:

• a + ar + ar^2 = 4.88
• a / 1 - r = 10 \(\displaystyle \ \ \ \ \ \ \ \) That should be a/(1 - r) = 10

So I thought that initially I would substitute the second equation with 'a' as the subject ( a = 10(1-r) ) into the first equation.

I tried this and it was obviously wrong (I was left with -10r^3 + 10 = 4.88). \(\displaystyle \ \ \ \ \ \ \ \) That is not wrong.

Could someone please tell me where I've gone wrong and how to solve for a and r?

\(\displaystyle -10r^3 + 10 \ = \ 4.88\)

\(\displaystyle -10r^3 \ = \ -5.12\)

\(\displaystyle 10r^3 \ = \ 5.12\)

\(\displaystyle r^3 \ = \ 0.512\)

\(\displaystyle r \ = \ \sqrt[3]{0.512}\)


Continue solving for r. \(\displaystyle \ \ \) Then use a = 10(1 - r) that you have above.

 

[Banks]

Ahh, I see, well that's good to know that I was actually on the right track. So r = 0.8 and a = 2.

Thanks for the help!