1. ## Geometric progression

Hello! I am quite confused about the formula needed for geometric progression, and would appreciate it if someone could explain how to solve geometric progression questions using the example question below...thank you!!

A geometric series has first term a and common ration r, where |r| < 1. The sum to infinity of the series is 8. The sum to infinity of the series is obtained by adding all the odd-numbered terms (i.e. 1st term + 3rd term + 5th term + ...) is 6. Find the value of r.

2. Originally Posted by Tangera
Hello! I am quite confused about the formula needed for geometric progression, and would appreciate it if someone could explain how to solve geometric progression questions using the example question below...thank you!!

A geometric series has first term a and common ration r, where |r| < 1. The sum to infinity of the series is 8. The sum to infinity of the series is obtained by adding all the odd-numbered terms (i.e. 1st term + 3rd term + 5th term + ...) is 6. Find the value of r.
Ok, the tool you need is the formula for a geometric series, which is $\displaystyle \frac{a}{1-r}$

For the first series you have that $\displaystyle \frac{a}{1-r}=8$ and for the second series you have that $\displaystyle \frac{a}{1-r^2}=6$; then you need to solve that system of equations for a and r.

3. Hello,

If |r|<1, then $\displaystyle \lim_{x \to \infty} |r|^x=0$

As the sum of a geometric series is $\displaystyle u_1 \frac{r^n-1}{r-1}$, the sum to infinity will be $\displaystyle u_1 \frac{1}{1-r}$

So $\displaystyle u_1 \frac{1}{1-r}=8$

If you skip the even-numbered terms, the new ratio will be r².

So $\displaystyle u_1 \frac{1}{1-r^2}=6$

I think you have pretty good things to conclude

4. Originally Posted by Tangera
A geometric series has first term a and common ration r, where |r| < 1. The sum to infinity of the series is 8. The sum to infinity of the series is obtained by adding all the odd-numbered terms (i.e. 1st term + 3rd term + 5th term + ...) is 6. Find the value of r.
Hello Tangera

so we have our geometric series.

$\displaystyle a + ar + ar^2 + ar^3 \cdots = 8$

use the formula for the geometric progression to simplify the LHS.

$\displaystyle \frac{a}{1-r} = 8 \ \ \ (1)$

were also given that

$\displaystyle a + ar^2 + ar^4 + ar^6 \cdots = 6$

This is another geometric series now with the common ratio of $\displaystyle r^2$

so it will sum to

$\displaystyle \frac{a}{1-r^2} = 6$
$\displaystyle \Rightarrow \frac{a}{(1-r)(1+r)} = 6 \ \ \ (2)$

so need to solve the simultaneous equation.

$\displaystyle \frac{a}{1-r} = 8 \ \ \ (1)$

$\displaystyle \frac{a}{(1-r)(1+r)} = 6 \ \ \ (2)$

notice that

$\displaystyle \frac{(1)}{(1+r)} = (2)$

$\displaystyle \frac{8}{(1+r)} = 6$

Now solve for r and a.

Bobak

5. Not to beat a dead horse, but since each term is modified by the ratio r, if you were to take every third term (1st, 4th, 7th, etc.), the ratio for that series would be $\displaystyle r^3$. Now suppose instead that you were taking all of the even-numbered terms. The first term of that series would be $\displaystyle ar$, and the sum to infinity would be $\displaystyle \frac{ar}{1-r^2}$.

6. I don't agree... It depends on what you define as being the first term. Since you work on infinity, you don't care about adding one or two terms, do you ?

Pardon me, I see the mistake now
I'm not sure, I just hope I understood correctly

7. Hello!

Thank you very very much to everyone who has replied! I was really glad you all explained how to get the a/(1-r^2) part of the solution....Thank you!! I have a clearer picture of how to solve such questions now. [aik! but I am afraid I might have more questions...my tutorial is difficult to me. sigh]