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

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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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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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Difficulty: 600
[Reveal] Spoiler: OA

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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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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.

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, 05:11
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] New post 14 Jun 2012, 19:27
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] New post 14 Jun 2012, 20:00
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] New post 14 Jun 2012, 22:36
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] New post 14 Jun 2012, 22:47
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] New post 16 Jun 2012, 02:45
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

Answer: E
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post 22 Jun 2012, 01:38
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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.

Answer: E.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post 19 Dec 2012, 21:38
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

Answer: (E) 28
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post 14 Jul 2014, 19:56
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Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1 [#permalink] New post 27 Aug 2014, 16:26
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.
Re: The sequence a1, a2, a3, ... ,an, ... is such that an = an-1   [#permalink] 27 Aug 2014, 16:26
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