Which of the following could be true of at least some of the terms of

Hey chetan2u, you're right. I misread that n was an integer, my brain did all the rest

the answer is E
where N could be any positive number, negative number or zero

MAGOOSH OFFICIAL SOLUTION:

First of all, if n = 8, then \(b_8 = (16 - 1)(16 + 3) = 15*19\)

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then \(b_{12} = (24 - 1)(24 + 3) = 23*27\)

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = (D).

Hi Magoosh Team ,
The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at \(N= \frac{1}{2}\) 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks
lucky

\(b_n\) is n-th term in the sequence, so n must be an integer.

hi lucky,
although i too thought the same but i think since bn is a term in a sequence so n has to be an integer

\(b_n\) is n-th term in the sequence, so n must be an integer.[/quote]


i see. thanks Bunuel .

yeah , we have to be very careful in reading the question.

Hi All,

Sequence questions are always based on a pattern of some kind, but there might be more to the pattern than you might initially realize. It often helps to "map out" the first few terms in a sequence question, so you can "see" the numbers and spot any hidden patterns involved.

Here, we're given the 'equation' for the sequence: bN = (2N-1)(2N+3) where N is the Nth term in the sequence....

1st term = (2-1)(2+3) = (1)(5) = 5
2nd term = (4-1)(4+3) = (3)(7) = 21
3rd term = (6-1)(6+3) = (5)(9) = 45
4th term = (8-1)(8+3) = (7)(11) = 77
Etc.

From this, we can deduce a number of details:
1) EVERY term will be an ODD INTEGER
2) The sequence of "products" will work through EVERY positive odd integer (notice the 1, the 3, the 5, the 7, etc).

This helps us to prove that Roman Numerals I and III COULD be true and that II is false.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

since it's a COULD be and not a MUST be true question...we can find ways for I and III to work.
look at the sequence..we have 2 odd numbers multiplied. since neither of them has a factor of 2, we can eliminate choices with II right away - B, C, and E - out.

now..
for (2n-1)(2n+3), we can find values so that it would be divisible by 15 and by 27
suppose 2n+3=15. 2n=12. n=6
(12-1)(12+3)=11*15 - so yes, possible

suppose 2n+3=27
2n=24
n=12
(24-1)(24+3)=23x27 - so yes, possible.

D

Could n be a negative integer? May not be, I think.

n is an index number which indicates the position of the term in the sequence, it cannot be negative.

even-1=15, even=16 (possible)
even-1=18, even=19; or even+3=18, even=15 (not possible)
even-1=27, even=28 (possible)

Ans (D)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.