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April 6th, 2014, 11:52 PM   #1
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Geometric Series

Hi everyone,
And thank you for the assistance last time asked for help.
I have a question posed to me by my tutor. Unfortunately, the tutor is away for 2 weeks so I have no one to ask for direction.
The question is
"For the Geometric progression with T3 = 3 and T10 = 6561, determine the first term, common ratio, and the sum of the first 20 terms."
I am using the following formula........... an = a1 + (n - 1)d

3 = a1 + (3 - 1)d 6561 = a1 + (10 - 1)d
a1 = 3 - (3 - 1)d a1 = 6561 - (10 - 1)d
3 - (3 - 1)d = 6561 - (10 - 1)d
6561 - 3 = (10 - 1)d - (3 - 1)d
6558 = 10d - d - 3d + d
6558 = 7d
d = 6558/7
d = 936.86
For T3 = 3
3 = a1 + (3 - 1)936.86
3 = a1 + 2 x 936.86
a1 = 3 - 1873.72
a1 = -1870.72
I am not asking for answers......... just want to know if I am on solid foundations.

Thanks
Phil
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April 7th, 2014, 03:03 AM   #2
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You say you are working with a geometric series but you are using the formula for an arithmetic series!

nth term of a geometric series (with a as the first term and r as the common ratio): $a\cdot r^{n-1}$

So we have $a\cdot r^2=3$ and $a\cdot r^9=6561$. Solve for a in terms of r in the first and substitute that into the second to find r.

Now use the formula for the sum of a geometric series with $n=20$: $$S=\frac{a(1-r^n)}{1-r}$$

Last edited by greg1313; April 7th, 2014 at 03:16 AM.
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