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S96-09

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S96-09  [#permalink]

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New post 16 Sep 2014, 00:50
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Question Stats:

57% (00:53) correct 43% (00:49) wrong based on 79 sessions

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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

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Re S96-09  [#permalink]

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New post 16 Sep 2014, 00:50
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: S96-09  [#permalink]

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New post 26 Oct 2017, 07:51
Another method would be to

1. substitute an actual geometric series ex: a,b,c,d as 2,4,6,8
2. give any easy value to k. I chose k =2
3. Substitute these values back in to the answer options and verify

Hope this helps!
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Re: S96-09  [#permalink]

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New post 05 Dec 2017, 13:56
vishkatti2005 wrote:
Another method would be to

1. substitute an actual geometric series ex: a,b,c,d as 2,4,6,8
2. give any easy value to k. I chose k =2
3. Substitute these values back in to the answer options and verify

Hope this helps!


I also solved this by substitution / picking numbers, however, 2, 4, 6 and 8 is NOT a geometric sequence as 8/6 does not equal 6/4 does not equal 4/2

I tried 1 2 4 8 and k=2. For the last one (III.) I got 16 16 16 16, so just to make sure, I also tried k=3, which gave 81 54 36 (constant or r=3/2).
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Re: S96-09  [#permalink]

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New post 19 Jun 2018, 17:18
im gonna fail this stupid **** exam.... are these considered as hard questions?
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Re: S96-09  [#permalink]

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New post 19 Jun 2018, 20:14
NickCat wrote:
im gonna fail this stupid **** exam.... are these considered as hard questions?


Yes, this is a 700-Level Questions, so hard. You can start practicing easier questions from our questions bank: https://gmatclub.com/forum/search.php?view=search_tags

Also, check below topics for more on Quantitative section:

ALL YOU NEED FOR QUANT ! ! !

Ultimate GMAT Quantitative Megathread.

Hope it helps.
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S96-09  [#permalink]

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New post 18 Nov 2018, 09:02
This question gives clues by looking at the answers. My gut feeling tells me that the expression in I and III look similar to each other. The expression in II is outstanding with "plus" sign. When you are looking a sequence with a constant ratio like this, it's a good thinking to consider "multiplication" rather than "Addition". It was my thought proces while guessing the answers :) Very throughout explanation though.
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Re: S96-09 &nbs [#permalink] 18 Nov 2018, 09:02
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