# Complicated Geometric Series Question.

• Apr 10th 2009, 06:48 AM
db5vry
Complicated Geometric Series Question.
The question is:

The sum of the first term and the second term of a geometric series is equal to twice the sum of the second term and third term of the series.
a] given that the common ratio is positive, find the value of the common ratio.
b] the sum to infinity of the series is 12. find, correct to two decimal places, the sum of the first eight terms of the series.

I will be able to do the part' b] OK, but the first part is troubling and I find it complicated.

I see that u1 + u2 = 2 (u2 + u3)
So [ u1 = $\displaystyle ar^{-1}$, u2 = $\displaystyle ar$] = 2 x [ u2 = $\displaystyle ar$, u3 = $\displaystyle ar^2$ ] ...{I think}

What happens from here?
• Apr 10th 2009, 07:02 AM
e^(i*pi)
Quote:

Originally Posted by db5vry
The question is:

The sum of the first term and the second term of a geometric series is equal to twice the sum of the second term and third term of the series.
a] given that the common ratio is positive, find the value of the common ratio.
b] the sum to infinity of the series is 12. find, correct to two decimal places, the sum of the first eight terms of the series.

I will be able to do the part' b] OK, but the first part is troubling and I find it complicated.

I see that u1 + u2 = 2 (u2 + u3)
So [ u1 = $\displaystyle ar^{-1}$, u2 = $\displaystyle ar$] = 2 x [ u2 = $\displaystyle ar$, u3 = $\displaystyle ar^2$ ] ...{I think}

What happens from here?

A) $\displaystyle U_1 = a$ by definition.

$\displaystyle a + ar = 2(ar + ar^2)$

Factor out: $\displaystyle a(1+r) = 2ar(1+r)$

As $\displaystyle a(1+r)$ is a common factor you can cancel to give $\displaystyle 2r = 1$ and so $\displaystyle r = \frac{1}{2}$

B)
$\displaystyle S_{infinity} = \frac{a}{1-r}$ which allows you to find a.

The sum of n terms is $\displaystyle S_n = \frac{a(1-r^n)}{1-r}$