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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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Math Expert
Joined: 02 Sep 2009
Posts: 50007
The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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02 Apr 2018, 01:07
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35% (medium)

Question Stats:

64% (02:53) correct 36% (02:35) wrong based on 47 sessions

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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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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
Math Expert
Joined: 02 Sep 2009
Posts: 50007
Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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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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02 Apr 2018, 05:36
1
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

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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