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

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Math Expert
Joined: 02 Sep 2009
Posts: 45367
The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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14 Jun 2012, 02:38
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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

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Math Expert
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Posts: 45367
Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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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:

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

$$a_6=\frac{a_{5}+a_{4}}{2}$$ --> $$a_6=\frac{20+36}{2}$$ --> $$a_5=28$$.

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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;
$$\frac{a_5-a_3}2=a_6-a_5$$
$$a_6=\frac{3a_5-a_3}2=\frac{3*20-4}2=28$$

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

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14 Jun 2012, 06:11
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a4=2*a6-a5;

=>a4=40-4=36; therefore,a6=(36+20)/2=28;

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

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14 Jun 2012, 20:27
a5 = (a3+a4) / 2
=> a4 = 2(a5) - a3
= 2 x 20 - 4 = 36

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

Option E is correct.

Difficulty level is 600.
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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14 Jun 2012, 21:00
Its E

from the given formula we can deduce A4 and then take average of A4 and A5 and there is your answer
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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14 Jun 2012, 23:36
its E.

a5=(a4+a3)/2
substituting you get a4=36

now a6=(a5+a4)/2
substitute to get a6 as 28
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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14 Jun 2012, 23:47
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]

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16 Jun 2012, 03:45
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

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

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19 Dec 2012, 22:38
$$a_3 = 4$$
$$a_4 = ?$$
$$a_5 = 20$$
$$a_6 = ?$$

$$a_n=\frac{a_{n-1} + a_{n-2}}{2}$$
$$20(2)=\frac{4 + a_4}{2}$$
$$40=4 + a_4$$
$$36 = a_4$$

$$a_6 = \frac{20 + 36}{2}=28$$

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

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26 Aug 2016, 04:16
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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13 Jan 2017, 14:33
a3 =4, a5 = 20
a5 = $$\frac{a4 + a3}{2}$$
20 =$$\frac{a4 + 4}{2}$$
40 - 4 = a4
a4 = 36
a6 = $$\frac{20 + 36}{2}$$
28
E
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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18 Apr 2017, 22:27
Option E

$$a_n=\frac{a_{n-1}+a_{n-2}}{2}$$

a$$_5 = 20, a_3 = 4$$

$$a_5=\frac{a_{4}+a_{3}}{2}$$
$$a_4 = 36$$

$$a_6=\frac{a_{5}+a_{4}}{2}$$ = 28
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink]

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30 Apr 2018, 18:03
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1   [#permalink] 30 Apr 2018, 18:03
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