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In a sequence, each term starting with the third onward is defined by

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In a sequence, each term starting with the third onward is defined by [#permalink]

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In a sequence, each term starting with the third onward is defined by the following formula \(a_n=a_{n-1}-a_{n-2}\), where n is an integer greater than 2. If the first term of that sequence is 3 and the second term is 4, what is the value of the 70th of that sequence?

A. −4
B. −3
C. −1
D. 3
E. 4
[Reveal] Spoiler: OA

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In a sequence, each term starting with the third onward is defined by [#permalink]

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Bunuel wrote:
In a sequence, each term starting with the third onward is defined by the following formula \(a_n=a_{n-1}-a_{n-2}\), where n is an integer greater than 2. If the first term of that sequence is 3 and the second term is 4, what is the value of the 70th of that sequence?

A. −4
B. −3
C. −1
D. 3
E. 4


Given \(an=a{n-1}-a{n-2}\)

a3 = a2 - a1 => 4-3 = 1.
a4 = -3
a5 = -4
a6 = -1
a7 = 3
a8 = 4
a9 = 1
a10 = -3

So pattern is 3,4 ,1,-3,-4,-1,3,4,1 and -3 - first term terms...

70th term is -3.

IMO option B.

Last edited by msk0657 on 05 Oct 2016, 09:33, edited 1 time in total.

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Re: In a sequence, each term starting with the third onward is defined by [#permalink]

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New post 05 Oct 2016, 08:12
Answer is B: -3.

Sequence is 1,-3,-4,-1,3,4
The sequence repeats every 6 digits. The 3rd term is 1, so we know that the number 3rd, 9th,15th,21th, 27th, 33th...69th is alwais 1.
Hence the 70th term is next one, -3

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Re: In a sequence, each term starting with the third onward is defined by [#permalink]

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asdasasdasasdasasdas wrote:
Answer is B: -3.

Sequence is 1,-3,-4,-1,3,4
The sequence repeats every 6 digits. The 3rd term is 1, so we know that the number 3rd, 9th,15th,21th, 27th, 33th...69th is alwais 1.
Hence the 70th term is next one, -3


Exactly..bro..missed this... I have taken 1 as the first number but ideally it is 3rd number and obviously...70th term will be -3...

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In a sequence, each term starting with the third onward is defined by [#permalink]

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New post 07 Oct 2017, 12:17
Bunuel wrote:
In a sequence, each term starting with the third onward is defined by the following formula \(a_n=a_{n-1}-a_{n-2}\), where n is an integer greater than 2. If the first term of that sequence is 3 and the second term is 4, what is the value of the 70th of that sequence?

A. −4
B. −3
C. −1
D. 3
E. 4

Recursive sequences, where each term is defined in terms of the previous term(s), typically have a useful pattern in test questions I have seen.

Each term has to be found by using previous terms. There is no easy shortcut. Start with what is given, and calculate successive terms until a pattern emerges.

Given: \(A_1 = 3\), \(A_2 = 4\)

\(A_{n} = A_{(n-1)} - A_{(n-2)}\)

\(A_1 = 3\)
\(A_2 = 4\)
\(A_3 = 1\)
\(A_4 = -3\)
\(A_5 = -4\)
\(A_6 = -1\)
\(A_7 = 3\) (\(A_1\) again)
\(A_8 = 4\) (\(A_2\) again; you can stop)
\(A_9 = 1\)
\(A_{10} = -3\)

The easy mistake here would be to decide that \(A_{70} = 3\) because 70 is a multiple of 7, and \(A_7 = 3\). Incorrect.

Instead: find the number of terms in the cycle ("cyclicity"). The pattern repeats every 6 terms. This sequence has a cyclicity of 6.

Divide 70 by 6 to see where \(A_{70}\) "falls" in the cycle.

70/6 has remainder of 4. The R=4 is the subscript number of the term whose value you need. If remainder is 4, Term\(_4\), or fourth term in the cycle of 6, has the same value as Term\(_{70}\).

Or: remainder = ·4· means \(A_{70} = A_{·4·}\)

\(A_{70} = A_4 = -3\)

Answer B

*\(A_3\) = \((A_2) - (A_1)\)
\(A_3\) = \((4) - (3) = 1\)
\(A_3 = 1\)

Kudos [?]: 251 [0], given: 543

In a sequence, each term starting with the third onward is defined by   [#permalink] 07 Oct 2017, 12:17
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