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The sequence a1, a2, a3, ... ,an, ... is such that an = an-1
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14 Jun 2012, 02:38
4
14
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Difficulty:
5% (low)
Question Stats:
83% (01:22) correct 17% (01:19) wrong based on 1550 sessions
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The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?
(A) 12 (B) 16 (C) 20 (D) 24 (E) 28
Diagnostic Test Question: 3 Page: 20 Difficulty: 600
The sequence a1, a2, a3, ... ,an, ... is such that an = an-1
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14 Jun 2012, 02:39
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SOLUTION
The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?
(A) 12 (B) 16 (C) 20 (D) 24 (E) 28
Since given that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\), then:
Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1
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14 Jun 2012, 05:58
1
1
Hi,
This one has be solved step by step, and there is chance of making mistakes.
Difficulty level: 600
\(a_n=\frac{a_{n-1}+a_{n-2}}2\) or a_n is the average of last two terms, thus, \(\frac{a_3+a_4}2=a_5\) \(\frac{a_4+a_5}2=a_6\) Subtracting these equations; \(\frac{a_5-a_3}2=a_6-a_5\) \(a_6=\frac{3a_5-a_3}2=\frac{3*20-4}2=28\)
Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1
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27 Aug 2014, 17:26
Bunuel wrote:
The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?
(A) 12 (B) 16 (C) 20 (D) 24 (E) 28
Diagnostic Test Question: 3 Page: 20 Difficulty: 600
Nice but easy problem. I think the idea is based on Fibonacci sequence.
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The sequence a1, a2, a3, ... ,an, ... is such that an = an-1
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14 Jul 2016, 19:47
Bunuel wrote:
The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?
(A) 12 (B) 16 (C) 20 (D) 24 (E) 28
Please note: this question in the 2017 version of the OG (page 20, #3, Quant Diagnostic Test) contains a typo. It should say \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\), but instead it says \(a_n=\frac{a_{n+1}+a_{n-2}}{2}\) for all \(n\geq{3}\). The answer explanation on page 46, however, lists the correct formula.
Yes, even the GMAC makes mistakes!
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83% (01:22) correct 
