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a, b, and c are the first three terms, in that order, in a sequence of

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a, b, and c are the first three terms, in that order, in a sequence of  [#permalink]

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New post 15 May 2020, 07:55
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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


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Re: a, b, and c are the first three terms, in that order, in a sequence of  [#permalink]

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New post 15 May 2020, 09:08
(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




Bunuel wrote:
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


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Re: a, b, and c are the first three terms, in that order, in a sequence of  [#permalink]

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New post 15 May 2020, 09:35
Bunuel wrote:
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


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
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Re: a, b, and c are the first three terms, in that order, in a sequence of  [#permalink]

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New post 15 May 2020, 18:35
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 ?

here,order is c<a<b<d [the median of these three terms is a, and b = 10. If a > c ]
let, common ratio = 1/r [T2/T1= 1/r]
T1=10r^2
T2=10r
T3=10
since, r^2 positive, T1 positive , so all the terms are positive .

|a + b + c| = 15
or,(1+r+r^2)= 3/2
or,3= 2+2r+2r^2
or,2r^2+2r-1=0
or, r^2+r-1/2=0
or,(r+1/2)^2 =3/4
or,r=√3/2-1/2 =0.732/2 = 0.366 =0.37
r cannot be negative, otherwise a>c will not hold.

the product of the first 4 terms of this sequence
(10r^2) *(10r)*(10)*(10/r)
=10000r^2
=1000*0.136
=1360

Bunuel
where am I wrong here?
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Re: a, b, and c are the first three terms, in that order, in a sequence of  [#permalink]

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New post 16 May 2020, 00:44
1. Correct sequence of geometric series is {a,b,c,d}; hence second term is 10 (not 3rd).
2. All the terms may or may not be positive. Since, b>a>c, few terms must be negative.

preetamsaha wrote:
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 ?

here,order is c<a<b<d [the median of these three terms is a, and b = 10. If a > c ]
let, common ratio = 1/r [T2/T1= 1/r]
T1=10r^2
T2=10r
T3=10
since, r^2 positive, T1 positive , so all the terms are positive .

|a + b + c| = 15
or,(1+r+r^2)= 3/2
or,3= 2+2r+2r^2
or,2r^2+2r-1=0
or, r^2+r-1/2=0
or,(r+1/2)^2 =3/4
or,r=√3/2-1/2 =0.732/2 = 0.366 =0.37
r cannot be negative, otherwise a>c will not hold.

the product of the first 4 terms of this sequence
(10r^2) *(10r)*(10)*(10/r)
=10000r^2
=1000*0.136
=1360

Bunuel
where am I wrong here?
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Re: a, b, and c are the first three terms, in that order, in a sequence of   [#permalink] 16 May 2020, 00:44

a, b, and c are the first three terms, in that order, in a sequence of

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