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12 Mar 2026, 22:33
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C
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Difficulty:
65%
(hard)
Question Stats:
63% (03:03) correct
37%
(02:52)
wrong
based on 81
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The sequence \(a_1\), \(a_2\), \(a_3\), ... is defined by
\(a_1 = 0\), \(a_2 = 6\)
\(a_n = a_{(n−1)} − a_{(n−2)} + 6\) for \(n ≥ 3\).
What is the sum of the first 49 terms of the sequence?
A. 270
B. 282
C. 288
D. 294
E. 300
\(a_1 = 0\), \(a_2 = 6\)
\(a_n = a_{(n−1)} − a_{(n−2)} + 6\) for \(n ≥ 3\).
What is the sum of the first 49 terms of the sequence?
A. 270
B. 282
C. 288
D. 294
E. 300
12 Mar 2026, 23:47
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kevincan
a1 = 0
a2 = 6
a3 = (6-0)+6 = 12
a4 = 12-6+6 = 12
a5 = 12 - 12 + 6 = 6
a6 = 6 - 12 + 6 = 0
a7 = 0-6+6 = 0
a8 = 0-0+6 = 6
Now we have a pattern.
It repeats in cycles of 6. The value sequence are (0,6,12,12,6,0).
The sum = 0+6+12+12+6+0 = 36.
49 terms is (8 * cycles of 6 + the last term).
The last term is the beginning of the sequence = 0.
Sum = 8*36 = 288.
Option C
25 Mar 2026, 13:43
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in such questions we need to find a repetitive pattern which can make the required summation easier to compute.
Given, every third term and further =(previous term - second than previous term)+6
so the sequence follows as: 0,6,(6-0+6)=12,(12-6+6)=12,(12-12+6)=6,(6-12+6)=0,(0-6+6)=0,
(0-0+6)=6,(6-0+6)=12,(12-6+6)=12........
here we can clearly observe that the pattern of the first 6 terms will repeat indefinitely, which adds up to 36.
sum of 49 terms=sum of 48 terms(total 8 patterns) + the first term=36*8+0=288
C
Given, every third term and further =(previous term - second than previous term)+6
so the sequence follows as: 0,6,(6-0+6)=12,(12-6+6)=12,(12-12+6)=6,(6-12+6)=0,(0-6+6)=0,
(0-0+6)=6,(6-0+6)=12,(12-6+6)=12........
here we can clearly observe that the pattern of the first 6 terms will repeat indefinitely, which adds up to 36.
sum of 49 terms=sum of 48 terms(total 8 patterns) + the first term=36*8+0=288
C
