A term an is called a “cusp� of a sequence . . . .

This topic has expert replies
Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pm
Thanked: 6 times
Followed by:5 members
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.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 am
Location: Grand Central / New York
Thanked: 470 times
Followed by:34 members

by Jay@ManhattanReview » Tue Dec 12, 2017 9:48 pm
Vincen 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.
We have a1 = k and a_n =−2(a_{n−1})/3

=> 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
_________________
Manhattan Review GMAT Prep

Locations: New Haven | Doha | Stockholm | Pretoria | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

Master | Next Rank: 500 Posts
Posts: 415
Joined: Thu Oct 15, 2009 11:52 am
Thanked: 27 times
Jay@ManhattanReview wrote:
Vincen 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.
We have a1 = k and a_n =−2(a_{n−1})/3

=> 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.
16K/81 I believe , but doesn't change the answer