Question 1119640
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Since obtaining the geometric sequence did not involve altering the first and third terms, the third term must be *[tex \Large r^2] times the first.  From this, we conclude that the common ratio is 3.


Since the given numbers are in the ratio 1:5:9, they can be expressed as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x,\ 5x,\ 9x]


From our analysis of the common ratio, we know that the geometric sequence must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a,\ 3a,\ 9a]


And we know that *[tex \Large 5x] must be 8 larger than *[tex \Large 3a] and that *[tex \Large x\ =\ a].  From this we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  5x\ =\ 3x\ +\  8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\ =\ 4]


so the geometric sequence must be


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4,\ 12,\, 36]


And then the original three numbers must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4,\ 20,\ 36]
								
								
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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