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A geometric sequence is one in which the ratio of any term after the f

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A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post 11 Nov 2014, 08:16
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Tough and Tricky questions: Sequences.



A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters \(a\), \(b\), \(c\), \(d\) represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of \(k\)?

I. \(dk\), \(ck\), \(bk\), \(ak\)

II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)

III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)


A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III

Kudos for a correct solution.

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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post 11 Nov 2014, 11:37
1
Bunuel wrote:

Tough and Tricky questions: Sequences.



A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters \(a\), \(b\), \(c\), \(d\) represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of \(k\)?

I. \(dk\), \(ck\), \(bk\), \(ak\)

II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)

III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)


A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III

Kudos for a correct solution.


Note that if a,b,c and d are in GP then... b^2=ac and c^2=bd

Option 1: \(dk\), \(ck\), \(bk\), \(ak\) so we have c^2k^2=dk*bk...Sufficient 1 is GP...Option C ruled out
Option 3 is better bet so let's do that before: b^2*k^6=ak^4*ck^2 or b^2*k^6=ack^6...St 2 is also GP.... Option A and B are out

Option 2: \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)

(b+2k)^2=(a+k)*(c+k) or b^2+4k^2+4k=ac+ak+ck+k^2 ---->3k^2 =(a+c-4)k... or k(3k-(a+c-4))=0 Clearly it does not meet criteria for GP..

Ans is D
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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post Updated on: 11 Nov 2014, 15:34
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1
Somehow I feel that i got tricked
First, a geometric sequence has to be on this form ax ,bx, cx... ( x is the common factor(ratio))
so 1) is definitely correct
3) dividing the sequence will make it easy to understand ak^4, bk^3 is a geometric sequence with K as a common factor. The same would be for the rest ck^2, dk^1
So 3) is correct
2) is a straight arithmetic sequence
hence D is the answer (unless i got tricked since the title is tricky questions :P)
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Originally posted by clipea12 on 11 Nov 2014, 14:53.
Last edited by clipea12 on 11 Nov 2014, 15:34, edited 1 time in total.
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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post 11 Nov 2014, 23:41
2
a, b, c, d are in GA, then

\(\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = Constant\)

\(b^2 = ac\)

\(c^2 = bd\)

Sequence II fails. I & III are sure winner

Answer = D
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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post 12 Nov 2014, 02:58
Official Solution:


A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters \(a\), \(b\), \(c\), \(d\) represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of \(k\)?

I. \(dk\), \(ck\), \(bk\), \(ak\)

II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)

III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)


A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III


Our first task is to ensure that we understand the definition of a geometric sequence. Let's use the sequence given to us: \(a\), \(b\), \(c\), \(d\). We are told that the ratio of any term after the first to the preceding term is a constant. In other words, \(\frac{b}{a} = \text{some constant}\), which is the same constant for the other ratios (\(\frac{c}{b}\) and \(\frac{d}{c}\)). Let's name that constant \(r\). Thus, we have the following:
\(\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r\)

By a series of substitutions, we can rewrite the sequence in terms of just \(a\) and \(r\):

\(a\), \(b\), \(c\), \(d\) is the same as \(a\), \(ar\), \(ar^2\), \(ar^3\)

Rewriting the sequence this way highlights the role of the constant ratio \(r\). That is, to move forward in the sequence one step, we just multiply by a constant factor \(r\). Rewriting also lets us substitute into the alternative sequences and watch what happens.

I. \(dk\), \(ck\), \(bk\), \(ak\)

This sequence is the same as \(ar^3k\), \(ar^2k\), \(ark\), \(ak\). To move forward in the sequence, we divide by \(r\). This is the same thing as multiplying by \(\frac{1}{r}\). Since this factor is constant throughout the sequence, the sequence is geometric.

II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)

This sequence is the same as \(a + k\), \(ar + 2k\), \(ar^2 + 3k\), \(ar^3 + 4k\). To move forward in this sequence, we cannot simply multiply by a constant expression. The presence of the plus sign means that we will not have a constant ratio between successive terms, and this sequence is not geometric.

III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)

This sequence is the same as \(ak^4\), \(ark^3\), \(ar^2k^2\), \(ar^3k\). To move forward in the sequence, we multiply by \(r\) and divide by \(k\). In other words, we multiply by \(\frac{r}{k}\), which is a constant factor. This sequence is therefore geometric.

Only sequences I and III are geometric.


Answer: D
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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