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# For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If

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Joined: 02 Sep 2009
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For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If  [#permalink]

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04 Apr 2018, 22:10
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15% (low)

Question Stats:

87% (01:26) correct 13% (01:10) wrong based on 66 sessions

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For the sequence $$a_1$$, $$a_2$$, $$a_3$$ ... $$a_n$$, $$an$$ is defined by $$a_n = a_{n−1} + 3$$. If $$a_1 = 0$$, then what is $$a_{15} − a_{12}$$?

A. 3
B. 6
C. 9
D. 12
E. 15

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Joined: 05 Feb 2016
Posts: 144
Location: India
Concentration: General Management, Marketing
WE: Information Technology (Computer Software)
Re: For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If  [#permalink]

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04 Apr 2018, 22:25
Bunuel wrote:
For the sequence $$a_1$$, $$a_2$$, $$a_3$$ ... $$a_n$$, $$an$$ is defined by $$a_n = a_{n−1} + 3$$. If $$a_1 = 0$$, then what is $$a_{15} − a_{12}$$?

A. 3
B. 6
C. 9
D. 12
E. 15

$$a_{15} = a_{1} + 3*14$$
$$a_{12} = a_{1} + 3*11$$

$$a_{15} − a_{12}=9$$
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Joined: 22 May 2016
Posts: 2040
For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If  [#permalink]

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05 Apr 2018, 05:53
Bunuel wrote:
For the sequence $$a_1$$, $$a_2$$, $$a_3$$ ... $$a_n$$, $$an$$ is defined by $$a_n = a_{n−1} + 3$$. If $$a_1 = 0$$, then what is $$a_{15} − a_{12}$$?

A. 3
B. 6
C. 9
D. 12
E. 15

The sequence is defined by
$$a_n = a_{n−1} + 3$$. If the way to solve is not apparent, list a few terms.

$$a_1 = 0$$
$$a_2 = a_1 + 3 = (0 + 3) = 3$$
$$a_3 = a_2 + 3 = 6$$
$$a_4 = a_3 + 3 = 9$$
$$a_5 = 12$$

There are a lot of ways to handle this pattern.
You can extrapolate, for example, from the number of multiples of 3 between three terms, or rewrite the definition.*

Extrapolate from $$a_1$$ to $$a_4$$ and $$a_2$$ to $$a_5$$:
The difference between any three terms is 9
From $$a_{12}$$ to $$a_{15}$$ there are 3 terms
The difference between those terms is 9

Rewrite the definition
Express the values above as [something] + added 3s;[something] is 0 = $$a_1$$

$$a_1 = 0$$
$$a_2 = 3 = (0+3)$$
$$a_3 = 6 = (0+3+3)$$
$$a_4 = 9 = (0)+(3+3+3)$$
$$a_5 = 12 = (0)+(3+3+3+3) = (a_1 + (4)(3))$$

Each term's multiple of 3 is one fewer than the subscript. Rewritten definition of the AP:
$$a_{n} = a_1 + (n-1)d$$
$$a_{n} = a_1 + (n-1)3$$

$$a_{12} = 0 + (11*3) = 33$$
$$a_{15} = 0 + (14*3) = 42$$
$$a_{15} - a_{12} = (42-33) = 9$$

*If the pattern does not strike, list terms $$a_1$$ to $$a_{15}$$
$$0,3,6,9,12,15,18,21,24,27,30,$$
$$(33 = a_{12}),36,39,(42 = a_{15})$$
$$a_{15} - a_{12} = (42 - 33) = 9$$

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Joined: 31 May 2017
Posts: 322
Re: For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If  [#permalink]

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05 Apr 2018, 20:26
1
The numbers are addition of 3 to each numbers from a2 to a15.

a12 = 33
a15 = 42
a15 - a12 = 42-33 = 9

Ans: C

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Re: For the sequence a1, a2, a3 ... an, an is defined by an = an−1 + 3. If &nbs [#permalink] 05 Apr 2018, 20:26
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