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There is a sequence such that an = a(n-2) + 12, where n is an integer

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There is a sequence such that an = a(n-2) + 12, where n is an integer  [#permalink]

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New post 15 Nov 2019, 02:35
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42% (03:13) correct 58% (02:15) wrong based on 30 sessions

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Re: There is a sequence such that an = a(n-2) + 12, where n is an integer  [#permalink]

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New post 15 Nov 2019, 05:58
Bunuel 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) \(a_2=23\)


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\(a_n=a_{n-2}+12\)

A bit of look into this equation to help understand the question better..
1) The equation will give 2 series... one for odd numbers..\(a_1, a_3, a_5\)... and one for even numbers..\(a_2, a_4, a_6\)...
2) We require to know one in odd series and one in even series to answer for sure.
3) If 417 does not belong to the series, you will have to check both series, hence C
4) But if 417 is in one of the series, then the answer will be series in which it is there, hence A or B

\(a_n=a_{n-2}+12\), where n is an integer greater 2.

(1) \(a_1=21\)
\(a_n=a_{n-2}+12........a_3=a_1+12=21+12....\)..
Hence the series will be 21, 21+12, 21+2*12... so 417-21 or 396 should be a multiple of 12
396 = 12*33..........417=21+33*12
Yes, 417 is in the sequence.

(2) \(a_2=23\)
We know the answer can be only A, B or C. So we can skip this and answer A.
But let us solve it..
\(a_n=a_{n-2}+12........a_4=a_2+12=23+12....\)..
Hence the series will be 23, 23+12, 23+2*12... so 417-23 or 394 should be a multiple of 12
394 = 12*32+10..........
No, 417 is not in the EVEN sequence, but it may be in ODD sequence. Hence, insufficient.

A
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Re: There is a sequence such that an = a(n-2) + 12, where n is an integer   [#permalink] 15 Nov 2019, 05:58
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