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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, 09:16
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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, 12:37
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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, 16:34
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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, 15:53.
Last edited by clipea12 on 11 Nov 2014, 16:34, 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 12 Nov 2014, 00:41
3
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, 03:58
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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Re: A geometric sequence is one in which the ratio of any term after the f  [#permalink]

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New post 20 Dec 2018, 14:14
Bunuel wrote:

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

Important: the fact that a,b,c,d is a given GP in that order guarantees, implicitly, that a,b, c and d are not zero. Think about that!

\(\left( {\rm{I}} \right)\,\,\,\,{{ck} \over {dk}} = {{bk} \over {ck}} = {{ak} \over {bk}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{c \over d} = {b \over c} = {a \over b}\,\,\,\,\,\,\left( { \Leftrightarrow \,\,\,\,\,\,{d \over c} = {c \over b} = {b \over a}} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a,b,c,d\,\,\,{\rm{GP}}\,\,\,:\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{refute}}\,\,\,\,\left( {\rm{C}} \right)\)

\(\left( {{\rm{II}}} \right)\,\,\,{\rm{Take}}\,\,{\rm{GP}}\,\,\left( {a,b,c,d} \right) = \left( {1,2,4,8} \right)\,\,{\rm{and}}\,\,k = 1\,\,:\,\,\,\,\,\,\left( {2,4,7,12} \right)\,\,\,{\rm{not}}\,\,{\rm{GP}}\,\,\,\,{\rm{:}}\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refute}}\,\,{\rm{also}}\,\,\left( {\rm{B}} \right),\left( {\rm{E}} \right)\)

\(\left( {{\rm{III}}} \right)\,\,{{b{k^3}} \over {a{k^4}}} = {{c{k^2}} \over {b{k^3}}} = {{dk} \over {c{k^2}}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{b \over {ak}} = {c \over {bk}} = {d \over {ck}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \matrix{
\,{b \over {ak}} = {c \over {bk}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{b \over a} = {c \over b}\,\,\,\,\,\, \Leftrightarrow \,\,\,{b \over c} = {a \over b}\,\,\,\,\,\, \hfill \cr
\,{c \over {bk}} = {d \over {ck}}\,\,\,\, \Leftrightarrow \,\,\,\,\,{c \over b} = {d \over c}\,\,\,\,\,\, \Leftrightarrow \,\,\,{c \over d} = {b \over c}\, \hfill \cr} \right.\,\,\, \Leftrightarrow \,\,\,\,\,a,b,c,d\,\,\,{\rm{GP}}\,\,\,:\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\)


The correct answer is therefore (D).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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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 27 Feb 2019, 10:24
2
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.


Recall that in a geometric sequence, the ratio of any term after the first to the preceding term is called the common ratio. We can let a, b, c and d be 1, 2, 4, and 8, respectively, (notice that the common ratio is 2) and k be 3. Now let’s analyze each Roman numeral.

I. dk, ck, bk, ak

dk = 8(3) = 24, ck = 4(3) = 12, bk = 2(3) = 6 and ak = 1(3) = 3

The sequence is 24, 12, 6, 3. This is a geometric sequence with a common ratio of 1/2.

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

a+k = 1 + 3 = 4, b+2k = 2 + 2(3) = 8, c+3k = 4 + 3(3) = 13 and d+4k = 8 + 4(3) = 20

The sequence is 4, 8, 13, 28. However, this is NOT a geometric sequence because we don’t have a nonzero constant as the common ratio.

III. ak^4, bk^3, ck^2, dk

ak^4 = 1(3)^4 = 81, bk^3 = 2(3)^3 = 54, ck^2 = 4(3)^2 = 36, dk = 8(3) = 24

The sequence is 81, 54, 36, 24. This is a geometric sequence with a common ratio of 2/3.

Answer: D
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Re: A geometric sequence is one in which the ratio of any term after the f   [#permalink] 27 Feb 2019, 10:24
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