# Math Help - geometric progression

1. ## geometric progression

A geometric progression has first term a where a‡0 and common ratio r where r‡1. the difference between the fourth and first term is equal to four times the difference between the third term and the second term.

i) show that r³ - 4r² + 4r -1 = 0
ii) find two possible values for the ratio of the geometric progression.
ii) for the value of r which the progression is convergent, prove that the sum to infinity is 1/2 a (1 + √5)

thanks

2. The nth term in a geometric sequence is given by

$a_{n}=a_{1}r^{n-1}$

For part a you have $a_{1}r^{3}-a_{1}=4(a_{1}r^{2}-a_{1}r)$

$a_{1}r^{3}-a_{1}=4a_{1}r^{2}-4a_{1}r$

$a_{1}r^{3}-4a_{1}r^{2}+4a_{1}r-a_{1}$

Factor out $a_{1}$:

$a_{1}(r^{3}-4r^{2}+4r-1)$

Solve the cubic to find the possible values of r.

3. Hello, Gracey!

A geometric progression has first term $a \neq 0$ and common ratio $r \neq 1$
The difference between $a_4\text{ and }a_1$ is equal to 4 times the difference between $a_3\text{ and }a_2$

(a) Show that: . $r^3 - 4r^2 + 4r - 1 \:=\:0$
The first four terms are: . $\begin{Bmatrix}a_1 &=& a \\ a_2 &=& ar \\ a_3 &=& ar^2 \\ a_4 &=&ar^3 \end{Bmatrix}$

We are told that: . $a_4 - a_1 \:=\:4\left(a_3-a_2\right)$

So we have . $ar^3 - a \:=\:4(ar^2-ar) \quad\Rightarrow\quad ar^3 - 4ar^2 + 4ar - a \:=\:0$

. . Divide by $a\!:\;\;\boxed{r^3 - 4r^2 + 4r - 1 \:=\:0}$

(b) Find two possible values for $r.$
Factor: . $(r-1)(r^2-3r+1) \:=\:0$

We have: . $r-1\:=\:\quad\Rightarrow\quad r \,=\,1\quad\hdots \text{ which is not allowed}$

. . and: . $r^2-3r+1\:=\:0\quad\Rightarrow\quad \boxed{r \:=\:\frac{3\pm\sqrt{5}}{2}}$

(c) For the value of $r$ which the progression is convergent,
prove that the sum to infinity is: . $\frac{1}{2}a\left(1 + \sqrt{5}\right)$
To be convergent, $r < 1.$ . Hence, we will use: . $r \;=\;\frac{3-\sqrt{5}}{2}$

Formula: . $S \;=\;\sum^{\infty}_{n=0} ar^n \;=\;\frac{a}{1-r}$

$\text{So we have: }\;S \;=\;\frac{a}{1 - \frac{3-\sqrt{5}}{2}} \;=\;\frac{2a}{\sqrt{5}-1}$

Rationalize: . $S \;=\;\frac{2a}{\sqrt{5}-1}\cdot\frac{\sqrt{5}+1}{\sqrt{5}+1} \;=\;\frac{2a(\sqrt{5}+1)}{5-1}\;=\;\frac{2a(1 + \sqrt{5})}{4}$

. . Therefore: . $\boxed{S \;=\;\frac{1}{2}a(1 + \sqrt{5})}$