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The sequence S is defined by Sn = Sn – 1 + Sn – 2 – 1 for ea
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25 Jul 2018, 17:27
25 Jul 2018, 17:27
Expert Reply
00:00
Question Stats:
76% (01:44) correct
23% (01:29) wrong based on 46 sessions
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The sequence S is defined by \(S_n = S_{n - 1} + S{n - 2} - 1\) for each integer n ≥ 3. If \(S_1 = 11\) and \(S_3 = 10\), what is the value of \(S_5\)?
(A) 0
(B) 9
(C) 10
(D) 18
(E) 19
(A) 0
(B) 9
(C) 10
(D) 18
(E) 19
Re: The sequence S is defined by Sn = Sn – 1 + Sn – 2 – 1 for ea
[#permalink]
11 Aug 2018, 17:00
11 Aug 2018, 17:00
Expert Reply
Explanation
The sequence \(S_n = S_{n - 1} + S_{n - 2} - 1\) can be read as “to get any term in sequence S, sum the two previous terms and subtract 1.”
The problem gives the first term and the third term and asks for the fifth term:
Within the sequence \(S_1\) to \(S_3\), the problem gives two values but not the middle one (\(S_2\)). What version of the formula would include those three terms?
\(S_3 = S_2 + S_1 - 1\)
\(10 = S_2 + (11) - 1\)
\(10 = S_2 + 10\)
\(0 = S_2\)
To get each subsequent term, sum the two previous terms and subtract 1. Thus, \(S_4 = 10 + 0 - 1 = 9\) and \(S_5 = 9 + 10 - 1 = 18\):
The sequence \(S_n = S_{n - 1} + S_{n - 2} - 1\) can be read as “to get any term in sequence S, sum the two previous terms and subtract 1.”
The problem gives the first term and the third term and asks for the fifth term:
| 11 | 10 | |||
| \(S_1\) | \(S_2\) | \(S_3\) | \(S_4\) | \(S_5\) |
Within the sequence \(S_1\) to \(S_3\), the problem gives two values but not the middle one (\(S_2\)). What version of the formula would include those three terms?
\(S_3 = S_2 + S_1 - 1\)
\(10 = S_2 + (11) - 1\)
\(10 = S_2 + 10\)
\(0 = S_2\)
| 11 | 0 | 10 | ||
| \(S_1\) | \(S_2\) | \(S_3\) | \(S_4\) | \(S_5\) |
To get each subsequent term, sum the two previous terms and subtract 1. Thus, \(S_4 = 10 + 0 - 1 = 9\) and \(S_5 = 9 + 10 - 1 = 18\):
| 11 | 0 | 10 | 9 | 18 |
| \(S_1\) | \(S_2\) | \(S_3\) | \(S_4\) | \(S_5\) |
gmatclubot
Re: The sequence S is defined by Sn = Sn – 1 + Sn – 2 – 1 for ea [#permalink]
11 Aug 2018, 17:00
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