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by srcc25anu » Mon Mar 28, 2011 5:55 am
In geometric series S, S1 = 2, S2 = 12, S3 = 72, ... . Is k a member of S?

(1) n is a member of S, and n = 36k

(2) k/6 is a member of S

OA - B
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by force5 » Mon Mar 28, 2011 12:50 pm
IMO B

common ratio is 6

if k/6 is a term in this sequence then k has to be a term too. since common ratio between the two terms then would be 6.
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by MAAJ » Mon Mar 28, 2011 2:05 pm
Yup. It must be (B)

S1 = 2 -> S1 = (1/3)*6^1
S2 = 12 -> S2 = (1/3)*6^2
S3 = 72 -> S3 = (1/3)*6^3

1) n is a member of S, and n = 36k

k = n/36 AND n could be 2,12,72 etc... not sufficient

2) k/6 is a member of S

If k/6 is a member of S, this means that k/6 is the number before K in the list of numbers. If we multiple by 6 we get K which must be in the list, because all numbers are equal to "previous number" * 6
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by rohu27 » Tue Mar 29, 2011 7:37 pm
for st A:
n=36k is a member of the series, so term before 36k would be 6k and the one before would be K.
am i missing somethign here?
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by GMATGuruNY » Thu Mar 31, 2011 3:03 am
srcc25anu wrote:In geometric series S, S1 = 2, S2 = 12, S3 = 72, ... . Is k a member of S?

(1) n is a member of S, and n = 36k

(2) k/6 is a member of S

OA - B
I received a PM asking me to comment.

Statement 1: n is a member of S, and n = 36k.
If n=2:
2 = 36k.
k = 2/36 = 1/18.
k is not a member of S.

If n=72:
72 = 36k.
k = 72/36 = 2.
k is a member of S.
Since in the first case k is not a member of S and in the second case k is a member of S, insufficient.

Statement 2: k/6 is a member of S.
k/6 = (member of S)
k = 6*(member of S)
Each term in S is 6 times the previous term in S.
So if we multiply any member of S by 6, the product will be the next term in S.
Thus, k is a member of S.
Sufficient.

The correct answer is B.
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by MAAJ » Fri Apr 01, 2011 5:14 am
You are right, that would be K, but that doesn't mean that K is a member of S.
rohu27 wrote:for st A:
n=36k is a member of the series, so term before 36k would be 6k and the one before would be K.
am i missing somethign here?
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