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For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all

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For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 08 Feb 2017, 01:40
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For the infinite sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\), \(a_{n+1}\), \(a_n=3*a_{n−1}\) for all n>1. If \(a_5−a_2=156\), what is \(a_1\)?

A. -1
B. 2
C. 3
D. 4
E. 8

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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 08 Feb 2017, 02:07
1
a5 = 3(a4) = 9(a3) = 27(a2) = 81(a1)

a5 - a2 = 81(a1) - 3(a1) = 156
78(a1) = 156
a1 = 2

Answer: B
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 08 Feb 2017, 02:35
Bunuel wrote:
For the infinite sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\), \(a_{n+1}\), \(a_n=3*a_{n−1}\) for all n>1. If \(a_5−a_2=156\), what is \(a_1\)?

A. -1
B. 2
C. 3
D. 4
E. 8


\(a_n=3*a_{n−1}\)
\(a_5-a_2 = 156\)
\(a_5 = 27*a_2\)
\(a_2 = 156/26 = 6 = 3*a_1\)
\(a_1 = 2\)

Hence Option B is correct.
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 09 Feb 2017, 14:40
Answer is B (a1= 2)

a5-a2=156
=> 3*a4-3*a1=156
=>a4-a1=52 ---- (1)

a4= 3*a3=3*3*a2=3*3*3*a1 ---- (2)

Solving (1) and (2), a1=2
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 10 Feb 2017, 02:50
Bunuel wrote:
For the infinite sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\), \(a_{n+1}\), \(a_n=3*a_{n−1}\) for all n>1. If \(a_5−a_2=156\), what is \(a_1\)?

A. -1
B. 2
C. 3
D. 4
E. 8


Option A can be eliminated as \(a1\) is negative so other elements in the series should also be increasingly negative. Hence the sum \(a_5−a_2\) should also be negative which is not the case here.

Option B) Writing down the elements with \(a1=2\) , 2,6,18,54,162. So 162-6 = 156.

Hence B

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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 13 Feb 2017, 08:38
2
1
Bunuel wrote:
For the infinite sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\), \(a_{n+1}\), \(a_n=3*a_{n−1}\) for all n>1. If \(a_5−a_2=156\), what is \(a_1\)?

A. -1
B. 2
C. 3
D. 4
E. 8


We are given that a(n) = 3 x a(n-1)

Thus:

a(2) = 3 x a(1)

a(3) = 3 x 3 x a(1) = 9a(1)

a(4) = 3 x 9 x a(1) = 27a(1)

a(5) = 3 x 27 x a(1) = 81a(1)

We are given a(5) - a(2) = 156. Thus:

81a(1) - 3a(1) = 156

78a(1) = 156

a(1) = 2

Answer: B
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 18 Apr 2017, 22:18
Option B

\(an = 3*an-1; n>1.\)

\(a5-a2 =156\)

\(3^4a1 - 3^1a1 = 156\)

\(a1 =2\)
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 19 Apr 2017, 11:21
An easier way to solve this problem is process elimination.

Let's start with A
If a1 = -1 then the sequences would be: -1, -3, -9, -27, -81
In this case a5 - a2 is not equal 156 so A is out

B
If a1 = 2 then the sequences would be: 2, 6, 18, 54, 162
162 - 5 = 156

Answer is B :-D
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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New post 19 Apr 2017, 12:22
Bunuel wrote:
For the infinite sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\), \(a_{n+1}\), \(a_n=3*a_{n−1}\) for all n>1. If \(a_5−a_2=156\), what is \(a_1\)?

A. -1
B. 2
C. 3
D. 4
E. 8


The series is in Geometric progression.

\(\frac{a_5}{a_2}=3^3\)

solving we get \(27a_2-a_2=156\) ==> \(a_2=6\)

\(a_2=3*a_1\) ==> \(a_1=2\)
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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all  [#permalink]

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Re: For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all &nbs [#permalink] 29 Jul 2018, 23:28
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