# Finding common ratio and more?

• Jan 3rd 2009, 03:28 PM
Jukixel
Finding common ratio and more?
I actually don't even know what topic this is. O.o"
It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
a) Calculate the value of the common ratio r.

b) If the fourth term is 16/9, calculate the sum of the infinite series.

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.
• Jan 3rd 2009, 03:45 PM
Jester
Quote:

Originally Posted by Jukixel
I actually don't even know what topic this is. O.o"
It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
a) Calculate the value of the common ratio r.

b) If the fourth term is 16/9, calculate the sum of the infinite series.

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.

Hints:

Let the series be $S = a + a r + a r^2 +a r^3 + a r^4 + \cdots$. You said the series has a finite answer so it converges and thus $-1 < r < 1$

For (a) consider $\frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}$

which factors

$\frac{a(1+r)(1+r^2)}{a(1+r)} = \frac{13}{9}$ from which you can find $r$.

Once you know $r$ you should be able to find $a$ (that's using the fact in (b)). Do you know the formula for the sum of a converging infinite geometric series?
• Jan 3rd 2009, 03:48 PM
Jukixel
No, don't think I know it.
• Jan 3rd 2009, 03:50 PM
Jester
Quote:

Originally Posted by Jukixel
No, don't think I know it.

If $-1 < r < 1$ then the sum of

$S = a + ar + ar^2 + ar^3 + \cdots$

is

$S_{\infty} = \frac{a}{1-r}$
• Jan 3rd 2009, 03:58 PM
skeeter
Quote:

The ratio of the sum of the first four terms to the sum of the first two terms is 13:9
a) Calculate the value of the common ratio r.
let r be the common ratio ...

$\frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}$

$\frac{1 + r + r^2 + r^3}{1 + r} = \frac{13}{9}$

$9 + 9r + 9r^2 + 9r^3 = 13 + 13r$

$9r^3 + 9r^2 - 4r - 4 = 0$

$9r^2(r + 1) - 4(r + 1) = 0$

$(r+1)(9r^2 - 4) = 0$

$(r+1)(3r+2)(3r-2) = 0$

$r = -1$ or $r = -\frac{2}{3}$ or $r = \frac{2}{3}$

Quote:

An infinite geometric series whose terms are positive has a finite sum.
$r = \frac{2}{3}$ is the only valid solution

Quote:

b) If the fourth term is 16/9, calculate the sum of the infinite series.
$ar^3 = \frac{16}{9}$

sum of a finite infinite geometric series is $S = \frac{a}{1-r}$

Quote:

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.
sum of the first n terms of a geometric series is $S_n = \frac{a(1-r^n)}{1-r}$
• Jan 4th 2009, 04:23 PM
Jukixel
Thanks!