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.
sequence
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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
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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I received a PM asking me to comment.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
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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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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