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Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of
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15 Dec 2020, 11:53
15 Dec 2020, 11:53
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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
A. 8,000
B. 16,000
C. 32,000
D. 40,000
E. 48,000
Re: Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of
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15 Dec 2020, 22:34
15 Dec 2020, 22:34
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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
A. 8,000
B. 16,000
C. 32,000
D. 40,000
E. 48,000
a, b and c are in a GP which means they are a, ar and ar^2 respectively. Median is a and a > c so a < b. In increasing order, it looks like b ,a , c
Now this is odd. GPs usually go in a single direction. If r > 1, then the terms keep increasing and if r lies between 0 and 1, the terms keep reducing. So something must be negative here to get positive-negative terms. We know that b = ar = 10 which is positive, so both a and r must be negative.
Since the sum of three terms is 15, it seems likely that we have all integers.
\(ar = 10 = -1*-10 = -2*-5 = -5*-2 = -10*-1\)
\(|a + ar + ar^2| = 15\)
This means that r must be a relatively small value else the sum might shoot up. Try a = -5, r = -2
\(|-5 + -5*-2 + -5*(-2)^2| = 15\)
Valid
So product of first 4 terms = \(a*ar*ar^2*ar^3 = (-5)^4 * (-2)^6 = 40,000\)
Answer (D)
Consider a, b, c in a G.P. such that |a + b + c| = 15. The median of
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16 Dec 2020, 00:07
16 Dec 2020, 00:07
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Expert Reply
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
A. 8,000
B. 16,000
C. 32,000
D. 40,000
E. 48,000
I believe "Consider a, b, c in a G.P. " means a, b and c ARE in GP. It could easily mean the GP is {x, b, a, c...} and a, b, c are three terms in the GP.
Also, the question should specify that a, b, c are first three terms of the GP.
Quote:
a, b, c are first three terms of 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.?
Logically
The product of the first four terms will be \(a* ar* ar^2* ar^3=a^4*r^6=(a^2r^3)^2\), that is a SQUARE.
Now, |a+b+c|=15 does point towards all the terms to be integer, and at least none of them is a square root or cube root.
So our answer should be something that can be written as a SQUARE of something.
We cannot have THREE 0s in a square. Only D possible => 40,000
D
Algebraically
a, b, c are in GP, and b>a>c => \(ar>a>ar^2\)
1) \(ar>a......ar-a>0.....a(r-1)>0\).
If a>0, r>1 OR if a<0, r<1.
2) \(a>ar^2......a-ar^2>0.....a(1-r^2)>0\).
If a>0, \(1>r^2\) OR if \(a<0, r^2>1\).
But when a>0, both r>1 and 1<r^2 are not possible. Thus case not possible
When a<0, r<1 and r^2>1. Both r<1 and r^2>1 are possible only when r<-1.
Hence a<0 and r<-1
Now ar=10=-2*-5=-5*-2, so a is -2 or -5
But \(|a+b+c|=15........|a+ar+ar^2|=|a(1+r+r^2)|=15=-3*-5\). so, a has to be a multiple of 3 or 5.
Thus, a=-5 and r=-2
We are looking for \(a*ar*ar^2*ar^3=a^4*r^6=(-5)^4*(-2)^6=4*10^4=40000\)
D
