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The sequence a1, a2, a3, ... ,an, ... is such that an = an-1

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Bunuel
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The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  14 Jun 2012, 01:38
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86% (01:48) correct 14% (01:47) wrong based on 2876 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
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Bunuel
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  14 Jun 2012, 01:39
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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:

\(a_5=\frac{a_{4}+a_{3}}{2}\) --> \(20=\frac{a_{4}+4}{2}\) --> \(a_4=36\);

\(a_6=\frac{a_{5}+a_{4}}{2}\) --> \(a_6=\frac{20+36}{2}\) --> \(a_6=28\).

Answer: E.
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  14 Jun 2012, 04:58
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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;
\(\frac{a_5-a_3}2=a_6-a_5\)
\(a_6=\frac{3a_5-a_3}2=\frac{3*20-4}2=28\)

Thus, Answer is (E).

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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  14 Jun 2012, 22:47
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I'm not sure how efficient my method was:

a3 = 4 = (a2 + a1) / 2
:. 8 = a2 + a1

a5 = 20 = (a4 + a3) / 2
:. 40 = a4 + a3 = a4 + 4
:. a4 = 36

a6 = ?
a6 = (a5 + a4) / 2
a6 = (20 + 36) / 2 = 28

(E) 28
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  16 Jun 2012, 02:45
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I consider it to be higher then 600 level due to function involved although very each one if one is able to decipher.

Calculate the value of a4 as a5 and a3 is given. Thus a5=a4+a3/2
20=(a4+4)/2 or 40-4=a4.

a6= 20+36/2= 28

Answer: E
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  14 Jul 2016, 18:47
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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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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  31 May 2021, 13:45
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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

Solution:

We see that:

a_5 = (a_4 + a_3) / 2

Since we are given the values of a_5 and a_3, we can solve for a_4:

20 = (a_4 + 4) / 2

40 = a_4 + 4

36 = a_4

With a_4 = 36, we can find the value of a_6 as follows:

a_6 = (a_5 + a_4)/2

a_6 = (20 + 36)/2

a_6 = 56/2 = 28

Answer: E
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post  20 Dec 2022, 08:13
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