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For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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16 Jul 2016, 09:08
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For the infinite sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ..., for all n > 1, \(a_n\) = \(a_{n−1}\) + 4 if n is odd and \(a_n\) = \(a_{n−1}\)  1 if n is even. What is the value of \(a_1\) if \(a_{34}\) = 68? A) 10 B) 11 C) 12 D) 13 E) 21
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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16 Jul 2016, 11:03
EBITDA wrote: For the infinite sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ..., for all n > 1, \(a_n\) = \(a_{n−1}\) + 4 if n is odd and \(a_n\) = \(a_{n−1}\)  1 if n is even. What is the value of \(a_1\) if \(a_{34}\) = 68?
A) 10 B) 11 C) 12 D) 13 E) 21 \(a_{34}\) = 68 = \(a_{33}\)  1 = \(a_{32}\)  1 +4 ... and so on So observing the trend, \(a_{34}\) = \(a_{1}\) +(32/2) * 4  (34/2) * (1) [ 32/2 to get all the even n (from 32 to 2) and 34/2 to get all the odd n (from 33 to 1) including 33 and 1] \(a_{1}\) = 6864+17 = 21 Hence, answer will be E.  P.S. Don't forget to give Kudos
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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16 Jul 2016, 15:06
\(\begin{align}&a_n = a_{n−1} + 4 &&\text{if n is odd}\\ &a_n = a_{n−1}  1 &&\text{if n is even}\\ &a_{34} = 68\\ \hline\\ &a_n = a_{n2} + 3 && \text{Combine both equations}\\ &a_{n2} = a_n  3&& \text{(Subtracting 3 will move you 2 steps closer to $a_0$)}\\\\ &\text{Aim to obtain $a_2$ (from $a_{34}$), this takes 32 steps, (16 steps of 2)}\\ &a_2 = 68  16 \times 3 = 68  48 = 20\\ &a_1 = a_2 + 1 = 21\\\\ &\text{Answer: (E) 21}\end{align}\)
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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17 Jul 2016, 02:52
Is there a quick way to solve this question, instead of manually counting the increment/decrements? I mean what if we had a100 given and we were supposed to calculate a1, how to approach the question in that case?
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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17 Jul 2016, 03:47
14101992 wrote: EBITDA wrote: For the infinite sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ..., for all n > 1, \(a_n\) = \(a_{n−1}\) + 4 if n is odd and \(a_n\) = \(a_{n−1}\)  1 if n is even. What is the value of \(a_1\) if \(a_{34}\) = 68?
A) 10 B) 11 C) 12 D) 13 E) 21 \(a_{34}\) = 68 = \(a_{33}\)  1 = \(a_{32}\)  1 +4 ... and so on So observing the trend, \(a_{34}\) = \(a_{1}\) +(32/2) * 4  (34/2) * (1) [ 32/2 to get all the even n (from 32 to 2) and 34/2 to get all the odd n (from 33 to 1) including 33 and 1] \(a_{1}\) = 6864+17 = 21 Hence, answer will be E.  P.S. Don't forget to give Kudos Thank you for your explanation. Nonetheless, I think that the bold part of your explanation should rather be: 32/2 to get all the odd n (from 33 to 1, including 33 but excluding 1, as you are already including \(a_{1}\) in your formula) and 34/2 to get all the even n (from 34 to 2, including both).



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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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07 Oct 2016, 07:36
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EBITDA wrote: For the infinite sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ..., for all n > 1, \(a_n\) = \(a_{n−1}\) + 4 if n is odd and \(a_n\) = \(a_{n−1}\)  1 if n is even. What is the value of \(a_1\) if \(a_{34}\) = 68?
A) 10 B) 11 C) 12 D) 13 E) 21 we have 34 consecutive numbers... 17 are even 17 are odd.. since we need to find a1 > then we must disregard first odd number. since odd means +4, then multiply 16 by 4. we get 64. since we have 17 even numbers, we subtract 17. so... a1+6417=68 a1+47=68 a1=21. E



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For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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16 Oct 2016, 13:07
Bunuel  is there a faster way to do it? TIA.



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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
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21 Dec 2017, 22:24
a34 = 68 = a33  1 a33 = 69 = a32 + 4 a32 = 65 = a31  1 a31 = 66 = a30 + 4 a30 = 62 = a29  1 a29 = 63 = a28 + 4 a28 = 59 = a27  1 . . . if you observe, a34, a32, a30, .....................a2 are in A.P with common diff = 3 , a2, a4 .... a34 => total 17 terms
so \(a34 = a2 + (17  1)d\) => \(68 = a2 + 16 * 3\) => \(a2 = 20\) a2 = a1  1 = 20 => a1 = 21 => (E)




Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all
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