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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] New post 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] New post 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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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink] New post 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] New post 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] New post 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, \(a_{35} = 11 + 306 = 317\).


Why did you add 11?
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink] New post 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, \(a_{35} = 11 + 306 = 317\).


Why did you add 11?


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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