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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

Expert replies
by Gmat_mission » Sat Jun 06, 2020 5:08 am

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Answers

A

B

C

D

E

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Difficulty

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

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT
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Source: — Problem Solving |

If a,b,c,d are in G.P
Then,
b/a = c/b = d/c

This condition is only satisfied by options I and III
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Gmat_mission wrote:
Sat Jun 06, 2020 5:08 am
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

[spoiler]OA=D[/spoiler]

Solution:

We can let a, b, c, d, and k be 1, 2, 4, 8, and 2, respectively (notice that 1, 2, 4, 8 form a geometric sequence with common ratio = 2). Now, let’s analyze each Roman numeral with the numbers we have for a, b, c, d, and k.

I. dk, ck, bk, ak = 16, 8, 4, 2

This is a geometric sequence with common ratio = 1/2.

II. a + k, b + 2k, c + 3k, d + 4k = 3, 6, 10, 16

This is not a geometric sequence since 6/3 ≠ 10/6 ≠ 16/10.

III. ak^4, bk^3, ck^2, dk = 16, 16, 16, 16

This is a geometric sequence with common ratio = 1.

Answer: D

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