Geometric Progression - find ith and r Hi great if you could please help me solve this:
Given:
$$a_0=12$$
$$a_5=24$$
$$a_{10} =48$$
thus ratio $$a_0:a_5 = 2$$
$n =$ number of terms in scale $= 5 (a_0, a_1, a_2, a_3, a_4) $
$a_0 =$ first term $= 12$
What is the equation to find $i$-th value $a_i$ , and the common ratio. 
Thanks
Jean-Lou
 A: Hints:  (HUGE Hints)
1) If $\frac {a_{k+1}}{a_{k}} = r$ then what is 
$\frac {a_5}{a_4}\frac {a_4}{a_3}\frac {a_3}{a_2}\frac {a_2}{a_1}\frac{a_1}{a_0}$?  And if $\frac {a_5}{a_0} = 2$ then what is $r$?
2) And if $a_{i} = r*{a_{i-1}} = r*(r*a_{i-2}) = r*(r*(r*a_{i-3})) = ...$
3) And if $a_i = f(i)$ be the formula in hint 2)  then $a_5 = 2a_0$ then $f(5) = 2\times f(0)$.  Can that tell you what $r$ is?
....
Do any two of the three hints.  Do all three if you need to.
A: $i$-th term of the GP can be given by:$$a_i=a_0\times{r}^i$$
now $r$ can be determined using the above formula:$$\ \ \ \ \ \ \ \ \ \ \  \ \ \ a_5=a_0\times r^5$$$$\ \ \ \ \ \ \ \ \Rightarrow 24=12\times r^5$$$$\ \ \ \ \ \ \ \ \Rightarrow r^5=\frac{24}{12}=2$$$$\Rightarrow r=2^{1/5} $$

Note that this formula for $i$-th term is not always going to be valid. In this case it is valid because the initial term of your GP is called $a_0$. Sometimes, the initial term of a GP is called $a_1$, in that case the formula would be:$$a_i=a_1\times r^{i-1}$$
