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A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha

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A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha [#permalink]

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A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for n > 2, and it has the starting values of \(s_1=2\) and \(s_2=3\). Find the value of \(s_6\).

A. 25
B. 32
C. 36
D. 93
E. 279


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[Reveal] Spoiler: OA

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Re: A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha [#permalink]

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New post 14 Mar 2015, 00:58
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Answer is E


we are given the pattern in which each term in the sequence could find by using the previous and the term before the previous term in the sequence.


so, for instance to find s3 we need to know s2 and also s1 . we are given s1 = 2 and s2 = 3 and we are ask to find s6 , so in order to calculate s6 , we need to know s5 as well as s4 and in the same way we need to know s3 to find s5 and s4 too.

so ,lets find s3 at the first step. we have a formula : s3 = (sn-1 -1 ) * ( sn-2) here : sn-1 for n=3 is s2 and sn-2 for n=3 is s1

so , we have : s3= ( s2-1) * (s1) = ( 3-1) *( 2 ) = 2*2=4

in the same way to find s4 we need to know s3 and s2 . so, we have : s4 = ( s3-1) * ( s2) = ( 4-1) * ( 3) = 3*3=9


For s5 : s5= (s4-1) * ( s3) = (9-1) *(4) = 8*4=32

And FINALLY: s6 = ( s5-1) *(s4) = (32-1) *(9) =31*9 =279 answer : E

With the best regards,

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Re: A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha [#permalink]

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New post 15 Mar 2015, 22:14
Bunuel wrote:
A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for n > 2, and it has the starting values of \(s_1=2\) and \(s_2=3\). Find the value of \(s_6\).

A. 25
B. 32
C. 36
D. 93
E. 279


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is a recursive sequence, and we have to find it term by term.
Attachment:
gpp-s_img11.png
gpp-s_img11.png [ 7.92 KiB | Viewed 2079 times ]


Answer = (E).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha [#permalink]

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New post 17 Feb 2017, 15:24
I just came across this question on Magoosh, but instead of giving s1 and s2, they gave s1 and s4.
s1=2
s4=9
All other details are identical.

How would you solve this now?

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A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha [#permalink]

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New post 20 Aug 2017, 17:55
funnypanks wrote:
I just came across this question on Magoosh, but instead of giving s1 and s2, they gave s1 and s4.
s1=2
s4=9
All other details are identical.

How would you solve this now?

Kudos for correct response!


I saw this variation as well, and was initially completely lost, but if you can assume that each term is a positive integer, you could also assume that for s4 to equal 9, it's likely it equals 9 because 3 was multiplied by 3. The only other option would be 9 * 1 which we can rule out because the sequence only increases in value, and s1 = 2 & s4 = 9, so s2 or s3 cannot = 1 or 9.

So now that we know the s4 equation is 3 * 3, we know s4-2 (aka s2) = 3, and s4-1 (aka s3) = 4, and we can solve the sequence as shown above for s6.

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A sequence is defined by s_n=(s_{n-1}-1)(s_{n-2}) for n > 2, and it ha   [#permalink] 20 Aug 2017, 17:55
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