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18 Jun 2019, 00:33
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E
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Difficulty:
75%
(hard)
Question Stats:
63% (03:18) correct
37%
(02:47)
wrong
based on 169
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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
A. 10
B. 11
C. 12
D. 13
E. 21
Archit3110
18 Jun 2019, 11:05
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Bunuel
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
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
\(a_3\) = \(a_2\)+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
