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# M23-07

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Math Expert
Joined: 02 Sep 2009
Posts: 44600

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16 Sep 2014, 01:18
00:00

Difficulty:

15% (low)

Question Stats:

86% (01:37) correct 14% (01:11) wrong based on 83 sessions

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The sequence $$a_1$$, $$a_2$$, $$a_3$$, ... , $$a_n$$, ... is such that $$a_n=a_{n-1}+a_{n-2}$$ for all $$n \ge 3$$. If $$a_1 = 3$$ and $$a_5 = 6$$, what is the value of $$a_2$$ ?

A. -3
B. -1
C. 0
D. 2
E. 3
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44600

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16 Sep 2014, 01:18
1
KUDOS
Expert's post
Official Solution:

The sequence $$a_1$$, $$a_2$$, $$a_3$$, ... , $$a_n$$, ... is such that $$a_n=a_{n-1}+a_{n-2}$$ for all $$n \ge 3$$. If $$a_1 = 3$$ and $$a_5 = 6$$, what is the value of $$a_2$$ ?

A. -3
B. -1
C. 0
D. 2
E. 3

Since given that $$a_n=a_{n-1}+a_{n-2}$$, then:

$$a_5=a_4+a_3=(a_3+a_2)+(a_2+a_1)=((a_2+a_1)+a_2)+(a_2+a_1)=3*a_2+2*a_1$$.

So, we have that $$a_5=3*a_2+2*a_1$$, or $$6=3*a_2+2*3$$, which gives $$a_2=0$$.

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Intern
Joined: 24 Feb 2017
Posts: 39

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27 Apr 2017, 22:43
i didn't get it !
how can you solve this?
Manager
Joined: 02 May 2016
Posts: 82
Location: India
Concentration: Entrepreneurship
GRE 1: 317 Q163 V154
WE: Information Technology (Computer Software)

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26 May 2017, 07:32
1
KUDOS
mkumar26 wrote:
i didn't get it !
how can you solve this?

Nth term = (N-1)th term + (n-2)th term

Just substitute the values:

a1=3
a2=x
a3=a2+a1 =x+3
a4=a3+a2 =x+3+x =2x+3
a5=a4+a3 =2x+3+x+3 = 3x+6

now a5=6, so 3x+6=6 ~ 3x=0 ~ x=0.
Hence, a2=x=0.

This is lengthy process though!
Intern
Joined: 02 Jun 2015
Posts: 11
Location: United States

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14 Mar 2018, 14:56
I believe the best way to solve this question is using "TEST THE ANSWERS" or "BACKSOLVING".

Let's use "C" because zero is an easy number to work with.

An= An-1 + An-2

Where A1=3 and A5=6

If A2 = 0

A3= A2 + A1
A3= 0 + 3
A3 = 3

A4= A3 + A2
A4 = 3 + 0
A4 = 3

A5= A4 + A3
A5= 3 + 3
A5=6 That's a match

I hope it helps!!
Thanks! Ale
Re: M23-07   [#permalink] 14 Mar 2018, 14:56
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# M23-07

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