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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)

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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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New post 02 Apr 2018, 01:07
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The infinite sequence \(a_1\), \(a_2\),… \(a_n\), … is such that \(a_1 = 4\), \(a_2 = -2\), \(a_3 = 6\), \(a_4 = -1\), and \(a_n = a_{n-4}\) for \(n > 4\). If \(T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}\), what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

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Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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New post 02 Apr 2018, 01:34
Bunuel wrote:
The infinite sequence \(a_1\), \(a_2\),… \(a_n\), … is such that \(a_1 = 4\), \(a_2 = -2\), \(a_3 = 6\), \(a_4 = -1\), and \(a_n = a_{n-4}\) for \(n > 4\). If \(T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}\), what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133



from 10 to 85 total 19 grps can be formed ,each grp 4 values

a10=a6=a2=-2
simlarly
a11=6,
a12=-1
a13=4
Total sum of one grp=7
Total sum for 19 grps=7*19=133

Option E
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Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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New post 02 Apr 2018, 02:56
Bunuel wrote:
The infinite sequence \(a_1\), \(a_2\),… \(a_n\), … is such that \(a_1 = 4\), \(a_2 = -2\), \(a_3 = 6\), \(a_4 = -1\), and \(a_n = a_{n-4}\) for \(n > 4\). If \(T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}\), what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133


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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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New post 02 Apr 2018, 05:36
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1
Bunuel wrote:
The infinite sequence \(a_1\), \(a_2\),… \(a_n\), … is such that \(a_1 = 4\), \(a_2 = -2\), \(a_3 = 6\), \(a_4 = -1\), and \(a_n = a_{n-4}\) for \(n > 4\). If \(T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}\), what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133


First figure out what the sequence looks like. Since \(a_n = a_{n-4}\), \(a_5 = a_1 = 4\), \(a_6 = a_2 = -2\), \(a_7 = a_3 = 6\) and so on...
4, -2, 6, -1, 4, -2, 6, -1, 4, -2, 6, -1, 4, ...

The first 4 terms are repeated. The sum of these terms is 4 - 2 + 6 - 1 = 7

From 10th to 85th terms, we have 85 - 10 + 1 = 76 terms
This means we will have 76/4 = 19 complete sets of 4 terms each so a total sum of 19*7 = 133

Answer (E)

Note that it doesn't matter with which term we start. Since we have 76 (a multiple of 4) total terms, we will get 19 complete sets. So if we start with -2, we will end with 4.
-2, 6, -1, 4, -2, 6, -1, ... 4, -2, 6, -1, 4
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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2) &nbs [#permalink] 02 Apr 2018, 05:36
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