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# The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1

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The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  30 Jul 2018, 10:15
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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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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  03 Aug 2018, 09:26
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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: 171

Kudos [?]: 2915 [0], given: 394

Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  12 Aug 2018, 07:03
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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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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  13 Oct 2018, 20:35
34×9+11=317
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  23 May 2020, 02:29
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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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]  23 May 2020, 23:22
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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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