A term an is called a "cusp" of a sequence if a_n is an integer but a_{n+1} is not an integer. If a_5 is a cusp of the sequence a1,a2,..., an,... in which a1=k and an=−2(a_{n−1})/3 for all n>1, then k could be equal to:
A. 3
B. 16
C. 108
D. 162
E. 243
The OA is D.
This is a hard question for me. I would be thankful if an expert helps me here.
A term an is called a “cusp� of a sequence . . . .
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We have a1 = k and a_n =−2(a_{n−1})/3Vincen wrote:A term a_n is called a "cusp" of a sequence if a_n is an integer, but a_{n+1} is not an integer. If a_5 is a cusp of the sequence a1, a2,..., an, ... in which a1 = k and a_n =−2(a_{n−1})/3 for all n>1, then k could be equal to:
A. 3
B. 16
C. 108
D. 162
E. 243
The OA is D.
This is a hard question for me. I would be thankful if an expert helps me here.
=> a_2 =−2(a_{2−1})/3 = -(2/3)a_1 = -(2/3)k
=> a_3 = (2/3)^2k = 4k/9
=> a_4 = -(2/3)^3k = -8k/27
=> a_5 = (2/3)^4k = 32k/81
=> a_6 = -(2/3)^5k = -64k/243
Since a_5 = 32k/81 is "Cusp," it is an integer, thus k must be a multiple of 81. Thus, the correct answer must be D or E.
Since a_5 = 32k/81 is "Cusp," a_6 = -64k/243 is NOT an integer, thus k must NOT be a multiple of 243. Thus, the correct answer is D.
The correct answer: D
Hope this helps!
-Jay
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16K/81 I believe , but doesn't change the answerJay@ManhattanReview wrote:We have a1 = k and a_n =−2(a_{n−1})/3Vincen wrote:A term a_n is called a "cusp" of a sequence if a_n is an integer, but a_{n+1} is not an integer. If a_5 is a cusp of the sequence a1, a2,..., an, ... in which a1 = k and a_n =−2(a_{n−1})/3 for all n>1, then k could be equal to:
A. 3
B. 16
C. 108
D. 162
E. 243
The OA is D.
This is a hard question for me. I would be thankful if an expert helps me here.
=> a_2 =−2(a_{2−1})/3 = -(2/3)a_1 = -(2/3)k
=> a_3 = (2/3)^2k = 4k/9
=> a_4 = -(2/3)^3k = -8k/27
=> a_5 = (2/3)^4k = 32k/81
=> a_6 = -(2/3)^5k = -64k/243
Since a_5 = 32k/81 is "Cusp," it is an integer, thus k must be a multiple of 81. Thus, the correct answer must be D or E.
Since a_5 = 32k/81 is "Cusp," a_6 = -64k/243 is NOT an integer, thus k must NOT be a multiple of 243. Thus, the correct answer is D.