Sequence
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cubicle_bound_misfit
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Stuart@KaplanGMAT
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This is a simple sequence in which each term is 2 terms before it plus 11.El Cucu wrote:If An=(An-2) +11 when n>2. Is 633 in the sequence?
1) A1=39
2) A2=43
Accordingly, if we know any even numbered term, we can calculate all the even numbered terms; if we know any odd numbered term, we can calculate all the odd numbered terms.
(1) A1 = 39.
Well, 633 - 39 = 594. 594 = 11*54. Therefore, 633 = A1 + a multiple of 11, which means that 633 is in the sequence: sufficient.
(2) A2 = 43.
Well, 633 - 43 = 590. 590 is NOT a multiple of 11. However, this only tells us that no even numbered term of the sequence = 633. Since we know nothing about the odd numbered terms, it's still possible that one of those could be 633. Maybe yes, maybe no: insufficient.
(1) is suff, (2) isn't: choose (A).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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Hi Stuart, impressive knowledge of Math!!Stuart Kovinsky wrote:This is a simple sequence in which each term is 2 terms before it plus 11.El Cucu wrote:If An=(An-2) +11 when n>2. Is 633 in the sequence?
1) A1=39
2) A2=43
Accordingly, if we know any even numbered term, we can calculate all the even numbered terms; if we know any odd numbered term, we can calculate all the odd numbered terms.
(1) A1 = 39.
Well, 633 - 39 = 594. 594 = 11*54. Therefore, 633 = A1 + a multiple of 11, which means that 633 is in the sequence: sufficient.
(2) A2 = 43.
Well, 633 - 43 = 590. 590 is NOT a multiple of 11. However, this only tells us that no even numbered term of the sequence = 633. Since we know nothing about the odd numbered terms, it's still possible that one of those could be 633. Maybe yes, maybe no: insufficient.
(1) is suff, (2) isn't: choose (A).
Some silly questions may be...
I. Why if we know that any even numbered term we can calculate all the even numbered term? I.e if we have A3=50 so we can calculate ALL the even terms? How if the rule given is An=(An-2) +11?
II. Why if 633-39=594 (which is the same as 633-A1) if it's multiple of 11 so 633 is in the series? I'm still thinking in the rule given an=(an-2)+11 but can't figure out when you put A1 with 633 how can you arrive to such conclusion?. Many tks in advance.
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Stuart@KaplanGMAT
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When I refer to even number terms, I'm referring to A2, A4, A6, A8, ... In other words, not the value of the term, but the place the term holds in the sequence.El Cucu wrote:
I. Why if we know that any even numbered term we can calculate all the even numbered term? I.e if we have A3=50 so we can calculate ALL the even terms? How if the rule given is An=(An-2) +11?
We can calculate all the even numbered terms from one even numbered term because the rule skips terms; each term is related to the term TWO before it.
So, if we know that A2 is 43, we know that A4 is 54, A6 is 65, A8 is 76 and so on. However, we know nothing about A3, A5, A7, ...
Since we know that each term is the term two before it plus 11, and we know that A1 = 39, we know that:II. Why if 633-39=594 (which is the same as 633-A1) if it's multiple of 11 so 633 is in the series? I'm still thinking in the rule given an=(an-2)+11 but can't figure out when you put A1 with 633 how can you arrive to such conclusion?. Many tks in advance.
A3 = 39 + 11
A5 = A3 + 11
A7 = A5 + 11
A9 = A7 + 11
and so on.
However, we could rewrite A5, A7 and A9 in terms of A1 as well:
A5 = 39 + 2(11)
A7 = 39 + 3(11)
A9 = 39 + 4(11)
following this pattern, we can see that each odd numbered term of the sequence is simply
39 + a multiple of 11,
and that 39 + every multiple of 11 will be in the sequence.
Since we can write 594 as 54(11), looking at the pattern above we know that:
39 + 54(11) will definitely be in the sequence.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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