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15 May 2020, 08:55
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B
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Difficulty:
95%
(hard)
Question Stats:
45% (02:59) correct
55%
(03:47)
wrong
based on 78
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a, b, and c are the first three terms, in that order, in a sequence of numbers where each term after the first is found by multiplying the previous one by a non-zero constant. |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 sequence ?
A. 48,000
B. 40,000
C. 32,000
D. 30,000
E. 8,000
Are You Up For the Challenge: 700 Level Questions
A. 48,000
B. 40,000
C. 32,000
D. 30,000
E. 8,000
Are You Up For the Challenge: 700 Level Questions
15 May 2020, 10:08
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(a,b,c,d)
\(b=a*r=10\)
\(c=a*r^2\)
\(d=a*r^3\)
a>c
\(a>ar^2\)
\(a(r^2-1)<0\)
1. If a>0, 0<r<1
2. If a<0, r<-1
\(|a+ar+ar^2|=15\)
Case 1- \(a(1+r+r^2)=15 \)
\(\frac{10}{r}(1+r+r^2)= 15\)
\(2+2r+2r^2 =3r\)
\(2r^2-r+2=0\)
GMAC doesn't like complex numbers.
(Reject this case)
Case 2-
\(a(1+r+r^2)=-15 \)
\(\frac{10}{r}(1+r+r^2)= -15\)
\(2+2r+2r^2 =-3r\)
\(2r^2+5r+2=0\)
\(2r^2+4r+r+2=0\)
\(2r(r+2)+1(r+2)=0\)
\((2r+1)(r+2)=0\)
\(r=-\frac{1}{2}\)(Rejected) r must be less than -1
r=-2
b=a*r=10
Product of first 4 terms=\( a*ar*ar^2*ar^3\) = \(a^4r^4 *r^2\)= \((ar)^4 *r^2\) = \(10^4*4 \)= 40000
\(b=a*r=10\)
\(c=a*r^2\)
\(d=a*r^3\)
a>c
\(a>ar^2\)
\(a(r^2-1)<0\)
1. If a>0, 0<r<1
2. If a<0, r<-1
\(|a+ar+ar^2|=15\)
Case 1- \(a(1+r+r^2)=15 \)
\(\frac{10}{r}(1+r+r^2)= 15\)
\(2+2r+2r^2 =3r\)
\(2r^2-r+2=0\)
GMAC doesn't like complex numbers.
(Reject this case)Case 2-
\(a(1+r+r^2)=-15 \)
\(\frac{10}{r}(1+r+r^2)= -15\)
\(2+2r+2r^2 =-3r\)
\(2r^2+5r+2=0\)
\(2r^2+4r+r+2=0\)
\(2r(r+2)+1(r+2)=0\)
\((2r+1)(r+2)=0\)
\(r=-\frac{1}{2}\)(Rejected) r must be less than -1
r=-2
b=a*r=10
Product of first 4 terms=\( a*ar*ar^2*ar^3\) = \(a^4r^4 *r^2\)= \((ar)^4 *r^2\) = \(10^4*4 \)= 40000
Bunuel
15 May 2020, 10:35
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Bunuel
Given: a, b, and c are the first three terms, in that order, in a sequence of numbers where each term after the first is found by multiplying the previous one by a non-zero constant. |a + b + c| = 15, the median of these three terms is a, and b = 10. a > c
Asked: What is the product of the first 4 terms of this sequence ?
Let r be the common ratio of the sequence.
b = ar = 10 ; a = 10/r
c = br = ar^2 = 10r
|a+b+c| = 15 = |10/r + 10 + 10r|; |1/r + 1 + r| = 1.5; r+1/r=-2.5; r = -2
Median of {a,b,c} is a ;
c<a<b; 10r<10/r<10
(a, b, c) = (-5,10,-20)
Product of first 4 terms = (-5)*10*(-20)*40 = 40000
IMO B
