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Difficulty:
15%
(low)
Question Stats:
83% (01:51) correct
17%
(02:47)
wrong
based on 239
sessions
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The infinite sequence \(a1, \ a2, \ …, \ an, \ …\) is such that \(a_1 = 3, \ a_2 = -1, \ a_3 = 6, a_4 = -2\), and \(a_n = a_{n-4}\) for \(n > 4\). What is the sum of the first 83 terms of the sequence?
A. 120
B. 124
C. 128
D. 132
E. 136
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A. 120
B. 124
C. 128
D. 132
E. 136
_________________
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Kinshook
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Joined: 03 Jun 2019
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Given: The infinite sequence \(a1, \ a2, \ …, \ an, \ …\) is such that \(a_1 = 3, \ a_2 = -1, \ a_3 = 6, a_4 = -2\), and \(a_n = a_{n-4}\) for \(n > 4\).
Asked: What is the sum of the first 83 terms of the sequence?
a_1 + a_2 + a_3 + a_ 4 = 3 - 1 + 6 - 2 = 6
a_5 + a_6 + a_7 + a_8 = a_1 + a_2 + a_3 + a_ 4 = 6
...
Sum of each successive 4 terms = 6
83 = 20*4 + 3
Sum of first 83 terms = sum of first 80 terms = 20*6 + 3 - 1 + 6 = 120 + 8 = 128
IMO C
Asked: What is the sum of the first 83 terms of the sequence?
a_1 + a_2 + a_3 + a_ 4 = 3 - 1 + 6 - 2 = 6
a_5 + a_6 + a_7 + a_8 = a_1 + a_2 + a_3 + a_ 4 = 6
...
Sum of each successive 4 terms = 6
83 = 20*4 + 3
Sum of first 83 terms = sum of first 80 terms = 20*6 + 3 - 1 + 6 = 120 + 8 = 128
IMO C
Posts: 57
I cant understand this question, can anyone give a detailed explanation
Bunuel
Since \(a_n = a_{n-4}\), then:
\(a_5 = a_1 = 3\)
\(a_6 = a_2 = -1\)
\(a_7 = a_3 = 6\)
\(a_8 = a_4 = -2\)
...
So, the sum of successive blocks of 4 terms is the same: {3, -1, 6, -2} {3, -1, 6, -2} {3, -1, 6, -2} ...
The sum of each block is 3 + (-1) + 6 + (-2) = 6.
Out of the first 83 terms, we have 20 full blocks of {3, -1, 6, -2}, each summing to 6, and then the next three terms are 3 + (-1) + 6 = 8. Therefore, the sum of 83 terms is 20 * 6 + 8 = 128.
Answer: C.
Hope it's clear.
harjas2222
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Posts: 57
t
Bunuel
Thank you man, appreciated
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