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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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08 Feb 2017, 00:40
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Difficulty:

35% (medium)

Question Stats:

78% (01:58) correct 22% (02:59) wrong based on 124 sessions

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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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08 Feb 2017, 01:07
1
a5 = 3(a4) = 9(a3) = 27(a2) = 81(a1)

a5 - a2 = 81(a1) - 3(a1) = 156
78(a1) = 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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08 Feb 2017, 01:35
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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09 Feb 2017, 13:40

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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10 Feb 2017, 01:50
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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13 Feb 2017, 07:38
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

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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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18 Apr 2017, 21:18
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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19 Apr 2017, 10:21
An easier way to solve this problem is process elimination.

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

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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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19 Apr 2017, 11:22
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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29 Jul 2018, 22:28
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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, 22:28
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