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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1  Question banks Downloads My Bookmarks Reviews Important topics
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Retired Moderator Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
Expert's post 00:00

Question Stats: 83% (01:11) correct 16% (01:35) wrong based on 42 sessions
The sequence $$a_1, a_2, a_3, ....,$$ an is defined by $$a_n = 9 + a_{n – 1}$$ for each integer n ≥ 2. If $$a_1 = 11$$, what is the value of $$a_{35}$$?

[Reveal] Spoiler: OA
317

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Sandy
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
1
KUDOS
We are looking to find the value of 35th integer.
We know the value of first term=11
There remains 34 more. Each term after 1st increases by 9 hence the 35th term is given by 11+ 34*9 = 317

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Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 175

Kudos [?]: 3036 , given: 394

Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
Expert's post
Explanation

Each term in the sequence is 9 greater than the previous term. To make this clear, write a few terms of the sequence: 11, 20, 29, 38, etc.

$$a_{35}$$ comes 34 terms after $$a_1$$ in the sequence. In other words, $$a_{35}$$ is $$34 \times 9 = 306$$ greater than $$a_1$$.

Thus, $$a_{35} = 11 + 306 = 317$$.
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Sandy
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Intern Joined: 06 Jul 2018
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
34×9+11=317
Senior Manager  Joined: 10 Feb 2020
Posts: 496
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Kudos [?]: 137 , given: 300

Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
sandy wrote:
Explanation

Each term in the sequence is 9 greater than the previous term. To make this clear, write a few terms of the sequence: 11, 20, 29, 38, etc.

$$a_{35}$$ comes 34 terms after $$a_1$$ in the sequence. In other words, $$a_{35}$$ is $$34 \times 9 = 306$$ greater than $$a_1$$.

Thus, $$a_{35} = 11 + 306 = 317$$.

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Ever Tried? Ever Failed? No Matter. Try Again. Fail Again. Fail Better!! Manager Joined: 09 Mar 2020
Posts: 164
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
1
KUDOS
Farina wrote:
sandy wrote:
Explanation

Each term in the sequence is 9 greater than the previous term. To make this clear, write a few terms of the sequence: 11, 20, 29, 38, etc.

$$a_{35}$$ comes 34 terms after $$a_1$$ in the sequence. In other words, $$a_{35}$$ is $$34 \times 9 = 306$$ greater than $$a_1$$.

Thus, $$a_{35} = 11 + 306 = 317$$.

The formula is Tn = a + (n-1)d, a = first term, n = term to find, d = difference between two consecutive terms (since 9 is added with every term, that is the difference)
Where T35 = 11 + (35-1) 9
T35 = 11 + (34) 9
T35 = 317 Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1   [#permalink] 23 May 2020, 23:22
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