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

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Bunuel
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For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink] New post   18 Jun 2019, 00:33
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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] New post   18 Jun 2019, 11:05
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Bunuel 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


use the given sequence formula
we get a33=69, a32= 65,a31=66,a30=62,a29=63,a28=59

its observed that sequence is going as pattern +1,-4 for each consecutive term
so at a1 would be 13 IMO D
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink] New post   08 Jul 2019, 11:18
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\(a_2\) = \(a_1\)-1
\(a_3\) = \(a_2\)+4 = \(a_1\)+3
\(a_4\) = \(a_3\) - 1 = \(a_1\)+ 2
\(a_5\) = \(a_4\)+ 4 = \(a_1\)+ 6

for every odd number, we can write
\(a_{2n+1}\) = \(a_1\)+ 3*n -------------- 1)

given,
\(a_{34}\) = 68 = \(a_{33}\)- 1
\(a_{33}\) = 69

from 1

\(a_{16*2+1}\) = \(a_1\) + 16*3 = 69
\(a_1\) = 69-48 = 21

Ans E
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink] New post   10 Jul 2019, 17:26
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Bunuel 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 can rewrite the formulas as:

a(n-1) = a(n) - 4 if n is odd [*] and a(n-1) = a(n) + 1 if n is even [**]

Since a(34) = 68,

a(33) = 69, a(32) = 65, a(31) = 66, a(30) = 62, a(29) = 63, a(28) = 59, and so on.

We see that each even numbered term is 3 less than its next larger even-numbered term, so we can create a new formula for the even-numbered terms as follows:

a(n) = 68 + 3(n - 34)/2 if n is even

So, a(2) = 68 + 3(2 - 34)/2 = 68 + 3(-32)/2 = 68 - 48 = 20. Since a(1) = a(2) + 1 (by formula [**]), we have a(1) = 21.

Answer: E
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink] New post   06 Nov 2021, 05:13
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Re: For the infinite sequence of numbers a1, a2, a3, ..., an, ..., for all [#permalink]
06 Nov 2021, 05:13
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