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geometric sequence

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Manager
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geometric sequence [#permalink] New post 17 Dec 2011, 07:01
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 5 sessions
v, w, x, y, z
A geometric sequence is a sequence in which each term after
the fi…rst is equal to the product of the preceding term and a
constant. If the list of numbers shown above is an geometric
sequence, which of the following must also be a geometric
sequence?
I. 2v, 2w, 2x, 2y, 2z
II. v + 2, w + 2, x + 2, y + 2, z + 2
III. \sqrt{v}, \sqrt{w}, \sqrt{x}, \sqrt{y}, \sqrt{z}

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
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Re: geometric sequence [#permalink] New post 17 Dec 2011, 11:14
My choice is E .Let me know the answer so that i can validate and explain...
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Manager
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Joined: 24 Nov 2010
Posts: 212
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Followers: 2

Kudos [?]: 32 [0], given: 7

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Re: geometric sequence [#permalink] New post 18 Dec 2011, 09:47
yeah, E is correct.

i) let's say orig sequence v, w, x, y, z = a, ba, b*ba, b*b*ba.. then 2v, 2w, 2x, 2y, 2z will be, 2a, 2ba, 2b*ba, 2*b*b*ba... every number is still a multiplication of the previous term by the constant - b

iii) again lets say v, w, x, y, z = a, ba, b*ba, b*b*ba.. then \sqrt{v}, \sqrt{w}, \sqrt{x}, \sqrt{y}, \sqrt{z} = \sqrt{a}, \sqrt{ba}, \sqrt{b*b*a}, \sqrt{b*b*ba}...

The first term in the sequence is now \sqrt{a}, and every term in the new sequence is still equal to the previous term multiplied by a constant - \sqrt{b}
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Re: geometric sequence [#permalink] New post 19 Dec 2011, 01:38
Yes,i took the same route and got E :)
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Re: geometric sequence [#permalink] New post 19 Dec 2011, 02:10
my ans is E

lets say series is 2,4,8,16,32 where common multiple is 2
let us multiply each term by 3, new series becomes
6,12,24,48,96 which is again a GP whose common multiple is 2

same case applies for square root


while adding 2 to each term will not generate any GP
Re: geometric sequence   [#permalink] 19 Dec 2011, 02:10
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