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# S96-09

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Math Expert
Joined: 02 Sep 2009
Posts: 49298

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16 Sep 2014, 01:50
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Difficulty:

45% (medium)

Question Stats:

57% (00:53) correct 43% (00:49) wrong based on 72 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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Math Expert
Joined: 02 Sep 2009
Posts: 49298

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16 Sep 2014, 01:50
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.

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Intern
Joined: 16 Oct 2017
Posts: 3
Location: India
WE: Business Development (Non-Profit and Government)

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26 Oct 2017, 08:51
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!
Intern
Joined: 16 Jul 2013
Posts: 33
Location: Hungary
Concentration: Entrepreneurship, Marketing
GMAT 1: 720 Q48 V40
GPA: 3.37

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05 Dec 2017, 14:56
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).
Intern
Joined: 16 Jun 2018
Posts: 3

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19 Jun 2018, 18:18
im gonna fail this stupid **** exam.... are these considered as hard questions?
Math Expert
Joined: 02 Sep 2009
Posts: 49298

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19 Jun 2018, 21:14
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 ! ! !

Hope it helps.
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Re: S96-09 &nbs [#permalink] 19 Jun 2018, 21:14
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# S96-09

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