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

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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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New post 21 Dec 2018, 02:56
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

91% (01:39) correct 9% (02:20) wrong based on 38 sessions

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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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New post Updated on: 21 Dec 2018, 08:49
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

Originally posted by Archit3110 on 21 Dec 2018, 03:52.
Last edited by Archit3110 on 21 Dec 2018, 08:49, edited 1 time in total.
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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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New post 21 Dec 2018, 05:25

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

Answer: E

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The sequence a1, a2, a3, …, an is defined by an = 9 + a(n – 1) for eac   [#permalink] 21 Dec 2018, 05:25
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