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M23-07

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M23-07  [#permalink]

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New post 16 Sep 2014, 01:18
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

85% (01:37) correct 15% (01:15) wrong based on 99 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

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Re M23-07  [#permalink]

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New post 16 Sep 2014, 01:18
1
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 \(a_2=0\).


Answer: C
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M23-07  [#permalink]

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New post 27 Apr 2017, 22:43
i didn't get it !
how can you solve this?
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Re: M23-07  [#permalink]

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New post 26 May 2017, 07:32
1
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!
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Re: M23-07  [#permalink]

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New post 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 &nbs [#permalink] 14 Mar 2018, 14:56
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