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

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Math Expert
Joined: 02 Sep 2009
Posts: 41875

Kudos [?]: 128460 [0], given: 12173

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

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05 Oct 2016, 03:28
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45% (medium)

Question Stats:

64% (02:26) correct 36% (02:28) wrong based on 74 sessions

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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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Kudos [?]: 128460 [0], given: 12173

Director
Joined: 26 Nov 2012
Posts: 593

Kudos [?]: 168 [0], given: 45

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

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05 Oct 2016, 04:19
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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.

Kudos [?]: 168 [0], given: 45

Intern
Joined: 29 Aug 2016
Posts: 6

Kudos [?]: 1 [0], given: 9

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

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05 Oct 2016, 08:12

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

Kudos [?]: 1 [0], given: 9

Director
Joined: 26 Nov 2012
Posts: 593

Kudos [?]: 168 [1], given: 45

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

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05 Oct 2016, 08:32
1
KUDOS
asdasasdasasdasasdas wrote:

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...

Kudos [?]: 168 [1], given: 45

Director
Joined: 22 May 2016
Posts: 797

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

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

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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 $$A_{70} = A_{·4·}$$

$$A_{70} = A_4 = -3$$

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

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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