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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
34×9+11=317
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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, \(a_{35} = 11 + 306 = 317\).


Why did you add 11?
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Re: The sequence a1, a2, a3, …, an is defined by an = 9 + an – 1 [#permalink]
1
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
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Re: The sequence a1, a2, a3, , an is defined by an = 9 + an 1 [#permalink]
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