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# There is a sequence such that

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Intern
Joined: 28 Dec 2014
Posts: 9
GMAT 1: 710 Q47 V41
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27 Jun 2017, 14:22
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95% (hard)

Question Stats:

43% (02:20) correct 57% (02:08) wrong based on 78 sessions

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There is a sequence such that $$a_n=a_{n-2}+12$$, where n is an integer greater 2. Is 417 contained in the sequence $$a_n$$?

1) $$a_1=21$$
2) $$a_2=23$$
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Joined: 02 Aug 2009
Posts: 6810
Re: There is a sequence such that  [#permalink]

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27 Jun 2017, 21:10
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aazt wrote:
There is a sequence such that $$a_n=a_{n-2}+12$$, where n is an integer greater 2. Is 417 contained in the sequence $$a_n$$?

1) $$a_1=21$$
2) $$a_2=23$$

Hi,

the sequence given will have two separate sequences, one of ODD values and other at EVEN values.
So for YES, if value of any one sequence has 417 it is sufficient
However for NO, we require both sequences.

Let's see the statements..
I. $$a_1=21$$..
Now $$a_3=21+12$$ and so on..
If 417 is in sequence, 417-21 should be divisible by 12.. 417-21=396, WHICH is div by 12
Suff
II. $$a_2=23$$
Working as above, we have answer NO for this even sequence.
But we do not know anything about the other sequence.
Insufficient

A
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There is a sequence such that  [#permalink]

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28 Jun 2017, 01:01
aazt wrote:
There is a sequence such that $$a_n=a_{n-2}+12$$, where n is an integer greater 2. Is 417 contained in the sequence $$a_n$$?

1) $$a_1=21$$
2) $$a_2=23$$

1) $$a_1=21$$

$$a_3 = a_1 + 12 = 21 + 12 = 33$$

Sequence is giving odd values.

$$33 - 21 = 12$$. 12 is divisible by 12.

To check if 417 is in the sequence.

$$417 - 21 = 396$$. 396 is divisible by 12.

Therefore the sequence will contain 417. Hence I is Sufficient.

2) $$a_2=23$$

$$a_2 = a_0 + 12 = 23$$

To check if 417 is in the sequence.

$$417 - 23 = 394$$. 394 is not divisible by 12.

Therefore this even sequence will not contain 417. However we are not given the odd sequence. Hence we cannot find if 417 is in the sequence. Hence II is Not Sufficient.

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Joined: 15 Jan 2017
Posts: 366
Re: There is a sequence such that  [#permalink]

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02 Oct 2017, 00:51
sashiim20 wrote:
aazt wrote:
There is a sequence such that $$a_n=a_{n-2}+12$$, where n is an integer greater 2. Is 417 contained in the sequence $$a_n$$?

1) $$a_1=21$$
2) $$a_2=23$$

1) $$a_1=21$$

$$a_3 = a_1 + 12 = 21 + 12 = 33$$

Sequence is giving odd values.

$$33 - 21 = 12$$. 12 is divisible by 12.

To check if 417 is in the sequence.

$$417 - 21 = 396$$. 396 is divisible by 12.
Have a query here: why are we minusing 21 instead of 12? I thought the 'd' which is 12 would be used.

Therefore the sequence will contain 417. Hence I is Sufficient.

2) $$a_2=23$$

$$a_2 = a_0 + 12 = 23$$

To check if 417 is in the sequence.

$$417 - 23 = 394$$. 394 is not divisible by 12.

Therefore this even sequence will not contain 417. However we are not given the odd sequence. Hence we cannot find if 417 is in the sequence. Hence II is Not Sufficient.

Senior Manager
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
Re: There is a sequence such that  [#permalink]

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03 Oct 2017, 02:10
There is a sequence such that $$a_n=a_{n−2}+12$$ where n is an integer greater 2. Is 417 contained in the sequence $$a_n$$?

=>$$a_n=a_{n−2}+12$$

=>$$a_3=a_1+12$$
=>$$a_4=a_2+12$$
=>$$a_5=a_3+12 = a_1+12*2$$
=>$$a_6=a_4+12 = a_2+12*2$$

We have to check if 417 is in the sequence or not

1) $$a_1$$=21
As we know $$a_1$$, we can find all the odd terms in the sequence

So there will be an odd term "p" in sequence such that => $$a_p=a_1+12*x = 21+12*x$$ Here x is an integer
417 can be written as 21 +12*33
hence 417 is present in the sequence
Sufficient

2) $$a_2$$=23
Same as above procedure : as we know $$a_2$$, we can find all the even terms in the sequence

So there will be an even term "q" in sequence such that => $$a_q=a_2+12*x = 23+12*x$$ Here x is an integer

417 cannot be written as 23 +12*x , where x is an integer
hence 417 is not present in the even position of the sequence.
But as we don't know anything about odd terms we cannot say if 417 is present in odd position in sequence or not
InSufficient

Re: There is a sequence such that &nbs [#permalink] 03 Oct 2017, 02:10
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