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# The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac

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Joined: 02 Sep 2009
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The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac  [#permalink]

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21 Dec 2018, 01:56
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Difficulty:

25% (medium)

Question Stats:

85% (01:36) correct 15% (01:45) wrong based on 26 sessions

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The sequence $$a_1$$, $$a_2$$, $$a_3$$, ...., $$a_n$$ is defined by $$a_n = 9 + a_{n – 1}$$ for each integer n ≥ 2. If $$a_1 = 11$$, what is the value of $$a_{35}$$?

A. 312
B. 313
C. 314
D. 316
E. 317

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The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac  [#permalink]

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Updated on: 21 Dec 2018, 07:49
a1= 11
d=11
an=a+(n-1)d

a35= 11+34*9
= 317 IMO e

Bunuel wrote:
The sequence $$a_1$$, $$a_2$$, $$a_3$$, ...., $$a_n$$ is defined by $$a_n = 9 + a_{n – 1}$$ for each integer n ≥ 2. If $$a_1 = 11$$, what is the value of $$a_{35}$$?

A. 312
B. 313
C. 314
D. 316
E. 317

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Originally posted by Archit3110 on 21 Dec 2018, 02:52.
Last edited by Archit3110 on 21 Dec 2018, 07:49, edited 1 time in total.
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Joined: 04 Jan 2015
Posts: 2456
The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac  [#permalink]

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21 Dec 2018, 04:25

Solution

Given:
• $$a_1, a_2, a_3$$ … are terms in a sequence
• $$a_n = 9 + a_{n – 1}$$, for n ≥ 2
• $$a_1 = 11$$

To find:
• The value of $$a_{35}$$

Approach and Working:
• $$a_2 = 9 + a_1$$
• $$a_3 = 9 + a_2 = 9 + 9 + a_3 = 9 * 2 + a_1$$
• $$a_4 = 9 + a_3 = 9 + 2 * 9 + a_1 = 3 * 9 + a_1$$
• Thus, we can infer that the terms are in AP, where the common difference is 9

Therefore, $$a_{35} = a_1 + (35 - 1) * 9 = 11 + 34 * 9 = 317$$

Hence, the correct answer is Option E

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The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac &nbs [#permalink] 21 Dec 2018, 04:25
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