1. Geomtric sequence question

Hey guys just started a new subject in maths today on geometric sequence

Any way here is the question I am having trouble with..

Find the value of x such that 3 - x; x; 2 - x are successive terms of a G.S and state the value of common ratio, r.

2. If $3-x\, ,x\, ,2-x$ are in GP, then

$x^2 = (3-x)(2-x)$
$\Rightarrow x^2 = 6 - 5x + x^2$
$\Rightarrow x = \frac65$

So the common ratio is $\frac{x}{3-x} = \frac{\frac65}{3-\frac65} = \frac{6}{15-6} = \frac23$

3. well geometric progression has the form of:
$ar^n$

the first term:[tex] ar^0 = a
the second term:[tex] ar^1 = ar
the third term: $ar^2 = ar^2$

to obtain r we divide a term with its previous term,

$\frac{ar^n}{ar^{n-1}} = r$

e.g.
$\frac{2ndterm}{1stterm} = \frac {ar}{a}$

and this is true through out the progression

4. $a_n = ..., ~3-x , ~x , ~2 - x,~...$

If this is a geometric sequence, every new value will be (previous . r).

$a_{n+1} = r\cdot a_n$

So, $2-x = x \cdot r$

$x = (3-x)\cdot r$

Then, we can write $r = \frac{2-x}{x} = \frac{x}{3-x}$

Now just solve it. Is it OK so far? This is the same as what isomorphism did, I only wanted to explain because I saw that you've just started this topic.

Now let's find a rule for these kinds of questions.

If $a$ is a term of a geometric sequence, the next terms will be $a\cdot r$ and $a \cdot r^2$.

$a,~a.r, ~a.r^2$

You can easily see that $(a)\cdot (a.r^2) = (a.r)^2$.

So, $a_{n-1} \cdot a_{n+1} = a_n^2$

Find the value of $x$ such that $3 - x,\; x,\; 2 - x$ are successive terms of a G.S
State the value of common ratio, $r.$
From the definition of the common ratio, we have: . $\begin{array}{cccc}\dfrac{x}{3-x} &=& r & {\color{blue}[1]} \\ \\ [-3mm] \dfrac{2-x}{x} &=& r & {\color{blue}[2]} \end{array}$

Equate [1] and [2]: . $\frac{x}{3-x} \:=\:\frac{2-x}{x}\quad\Rightarrow\quad\boxed{ x \:=\:\frac{6}{5}}$

Substitute into [2]: . $r \:=\:\frac{2-\frac{6}{5}}{\frac{6}{5}}\quad\Rightarrow\quad\box ed{ r \:=\:\frac{2}{3}}$