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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 \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... 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 [#permalink]

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14 Jun 2012, 05:58

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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 [#permalink]

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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.

The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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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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