Question 1119641
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The *[tex \Large n\text{th}] term of a geometric series is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_n\ =\ ar^{n-1}]


Where *[tex \Large r] is the common ratio:


So the first three terms are *[tex \Large ar^0,\ ar^1,\ ar^2] or more simply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a,\ ar,\ ar^2]


Since *[tex \Large ar^2] is given to be *[tex \Large 4a], we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ 4\ \Right\ r\ =\ 2]


And since *[tex \Large ar^2\ -\ ar\ =\ 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2a\ =\ 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 12]


So the first three terms are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12,\ 24,\ 48]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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