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Marriage of Arithmetic and Geometric Sequence

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Marriage of Arithmetic and Geometric Sequence

by dtweah » Tue May 19, 2009 12:31 pm
Let a, b, c, d be positive real numbers with a < b < c < d. Given that
a, b, c, d are the first four terms in an arithmetic sequence, and a, b, d are the first three terms in a geometric sequence, what is the value of
ad/bc ?

(a)1/2
(b)2/3
(c)3/4
(d)4/5
(e) 1
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by dumb.doofus » Tue May 19, 2009 2:37 pm
B i.e. 2/3

we can just choose a = 1, b = 2, c = 3 and d = 4

This satisfies both arithmetic and geometric series requirements as per the question..
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by El Cucu » Tue May 19, 2009 3:00 pm
Very interesting question!

My aproach:

a, b, c, d would equal in the arithmetic sequence:

a,a+x,a+2x, and a+3x

in the geometric sequence

a, b, d

a/b = b/d= y

so b=a*y and d=b*y^2

Then we have to make equations from the two sequences.
a=a
b=a+x=a*y
c=a+2x
d= a+3x=b*y^2

So a+3x= a*y^3

And here is when I need help from a Math Guru! May be by picking numbers/pluging in answers we can solve this!
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by dumb.doofus » Tue May 19, 2009 3:09 pm
El Cucu wrote:Very interesting question!

My aproach:

a, b, c, d would equal in the arithmetic sequence:

a,a+x,a+2x, and a+3x

in the geometric sequence

a, b, d

a/b = b/d= y

so b=a*y and d=b*y^2

Then we have to make equations from the two sequences.
a=a
b=a+x=a*y
c=a+2x
d= a+3x=b*y^2

So a+3x= a*y^3

And here is when I need help from a Math Guru! May be by picking numbers/pluging in answers we can solve this!
What's the point? After doing all this you still want to plugin values.. well, in that case, just plug in values like I have shown above... Wouldn't that be more simple?
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