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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
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\(a=\frac{10}{k}\)
c=10k

\(10k+10+\frac{10}{k} = 15\) or \(10k+10+\frac{10}{k} = -15\)

case 1- \(10k+10+\frac{10}{k} = 15\)

\(2k+2+\frac{2}{k} =3\)

\(2k^2 -k+2=0\)

We don't have real roots for this equation

case 2- \(10k+10+\frac{10}{k} = -15\)

\(2k+2+\frac{2}{k} =-3\)

\(2k^2 +5k+2=0\)

\((2k+1)(k+2)=0\)

k= -2 or \(-\frac{1}{2}\)

since a>c, reject k=\(-\frac{1}{2}\)

We get a=-5, b=10 and c=-20

Assuming that a is the first term of the sequence

Product of first 4 terms = -5*10*-20*40 = 40000






DisciplinedPrep wrote:
Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of these three terms is a, and b = 10. If a > c, what is the product of the first 4 terms of this G.P.?

A. 8,000
B. 16,000
C. 32,000
D. 40,000
E. 48,000
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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
Hi is GP supposed to mean geometric progression here?
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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
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CEdward wrote:
Hi is GP supposed to mean geometric progression here?


Yes, G.P. means geometric progression.
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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
Bunuel wrote:
CEdward wrote:
Hi is GP supposed to mean geometric progression here?


Yes, G.P. means geometric progression.



Hi Bunuel ,

I am not able to understand how we get that sequence should have a negative number ? Could you please help
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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
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Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of [#permalink]
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