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

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Kudos [?]: 146 [0], given: 33

For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]

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New post 16 Jul 2016, 08:08
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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
[Reveal] Spoiler: OA

Kudos [?]: 146 [0], given: 33

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

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New post 16 Jul 2016, 10:03
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]

\(a_{1}\) = 68-64+17 = 21

Hence, answer will be E.

-------------------------------------

P.S. Don't forget to give Kudos :)
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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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New post 16 Jul 2016, 14:06
\(\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_{n-2} + 3 && \text{Combine both equations}\\
&a_{n-2} = 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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New post 17 Jul 2016, 01:52
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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Kudos [?]: 1066 [0], given: 74

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Kudos [?]: 146 [0], given: 33

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

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New post 17 Jul 2016, 02:47
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]

\(a_{1}\) = 68-64+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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New post 07 Oct 2016, 06:36
1
This post was
BOOKMARKED
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+64-17=68
a1+47=68
a1=21.

E

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Kudos [?]: 51 [0], given: 284

For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]

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New post 16 Oct 2016, 12:07
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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New post 21 Dec 2017, 21:24
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)

Kudos [?]: 34 [0], given: 1146

Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all   [#permalink] 21 Dec 2017, 21:24
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