1. ## [SOLVED] Geometric Progression

Random question I came across when studying, couldn't work out how to do it.

Question:
The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

Thanks.

2. Originally Posted by aod
Random question I came across when studying, couldn't work out how to do it.

Question:
The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

Thanks.
You know that: $\displaystyle \frac{5x - 1}{x^2 - 1} = \frac{-3}{5x - 1}$

$\displaystyle (5x - 1)^2 = -3x^2 + 3$

$\displaystyle 25x^2 - 10x + 1 = -3x^2 + 3$

$\displaystyle 28x^2 - 10x - 2 = 0$

Solve for $\displaystyle x$, i'm lazy

(When you have x, you will have the terms and from there it will be easy to find the common ratio)

3. Originally Posted by aod
Random question I came across when studying, couldn't work out how to do it.

Question:
The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

Thanks.
recall that in a geometric progression, the nth term, $\displaystyle a_n$, is given by:

$\displaystyle a_n = ar^{n-1}$ for $\displaystyle n \in \mathbb{N} = 1,2,3,4,...$

thus, since the first term is $\displaystyle x^2 - 1$ it means that $\displaystyle a = x^2 - 1$

also, the second term is:

$\displaystyle a_2 = ar = \left( x^2 - 1 \right)r = 5x - 1$

$\displaystyle \Rightarrow r = \frac {5x - 1}{x^2 - 1}$ ...................(1)

the third term is:

$\displaystyle a_3 = ar^2 = \left( x^2 - 1 \right)r^2 = -3$

$\displaystyle \Rightarrow r = \sqrt{\frac {-3}{x^2 - 1}}$ ..................(2)

Now, equate (1) and (2), we get:

$\displaystyle \sqrt{\frac {-3}{x^2 - 1}} = \frac {5x - 1}{x^2 - 1}$

now solve that to et a range of values for $\displaystyle x$, and then you can find the possible values for $\displaystyle a$ and $\displaystyle r$

EDIT: Ah, Jan's way is easier

4. Ah makes sense now, thanks for help