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08 Aug 2017, 18:10
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85% (01:34) correct
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There is a sequence of r, t, s, 1, 4, 3, 8, 15, 26. Each number of the sequence is the sum of the 3 previous numbers. r=?
A. -2
B. -1
C. 0
D. 1
E. 2
A. -2
B. -1
C. 0
D. 1
E. 2
08 Aug 2017, 20:02
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MathRevolution
There is a sequence of r, t, s, 1, 4, 3, 8, 15, 26. Each number of the sequence is the sum of the 3 previous numbers. r=?
A. -2
B. -1
C. 0
D. 1
E. 2
A. -2
B. -1
C. 0
D. 1
E. 2
This sequence is possible when r = -2, s = 5, t = -2 as r+s+t = 1.
To double check whether the numbers we have selected are correct,
we also have s + t + 1 = 4 to prove that the numbers chosen are correct.
Hence, it is clear that Option A(r = -2) is our correct answer.
sashiim20
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Concentration: Technology, Strategy
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Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 612
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08 Aug 2017, 23:18
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MathRevolution
There is a sequence of r, t, s, 1, 4, 3, 8, 15, 26. Each number of the sequence is the sum of the 3 previous numbers. r=?
A. -2
B. -1
C. 0
D. 1
E. 2
A. -2
B. -1
C. 0
D. 1
E. 2
We can solve this question by solving from last right number.
Given Each number of the sequence is the sum of the \(3\) previous numbers.Therefore;
\(26 = (15 + 8 + 3) => 26\)
\(15 = (8 + 3 + 4) => 15\)
\(8 = (3 + 4 + 1) => 8\)
\(3 = (4 + 1 + s) => 5 + s\)
\(s = 3 - 5 => -2\)
\(4 = (1 + s + t)\)
\(4 = 1 - 2 + t => -1 + t\)
\(t = 4 + 1 => 5\)
\(1 = (s + t + r)\)
\(1 = -2 + 5 + r => 3 + r\)
\(r = 1 - 3 => -2\)
Answer (A)...
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