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In any sequence of n nonzero numbers, a pair of consecutive terms with
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26 Apr 2019, 02:32
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63% (01:39) correct 37% (01:56) wrong based on 110 sessions
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In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes? (1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd. DS57502.01 OG2020 NEW QUESTION
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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26 Apr 2019, 14:43
Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION Instead of reasoning about the sequence, we'll try simple numbers to give contradictory answers. This is an Alternative approach. (1) If n = 2 then the sequence is 1, 1 and we have 1 sign change. If n = 3 then the sequence is 1, 1, 1 and we have 2 sign changes. Insufficient. (2) With no information on the sequence we can build whatever we like... insufficient. Combined: If n = 3 we saw 2 sign changes. If n = 5 we have 1, 1, 1, 1, 1 which takes the same 2 sign changes from before and adds another 2. Similarly, each consecutive odd number will add another 2 sign changes so we will always have an even number of changes in total. (C) is our answer.
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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27 Apr 2019, 07:58
Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k=(–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION Let \(x\) be the number of sign changes of the sequence. The original question: Is \(x=E\) ? 1) The consecutive terms of the sequence defined in 1) will have opposite signs, so \(x=n1\). If \(n=E\), then \(x=EO=O\), so the answer to the original question is No. However, if \(n=O\), then \(x=OO=E\), so the answer to the original question is Yes. Thus, we can't get a definite answer to the original question. \(\implies\) Insufficient2) We know that the number of terms of the sequence is odd, but we don't know anything about the rule defining the consecutive terms. \(\implies\) Insufficient1&2) We know that \(n=O\), so \(x=OO=E\). Thus, the answer to the original question is a definite Yes. \(\implies\) SufficientAnswer: C
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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06 May 2019, 20:16
Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION s(k) = (1)^k We see that s(k) is the sequence: 1, 1, 1, 1. ... , (1)^n However, since we don’t know the value of n, we can’t determine the number of sign changes or whether it’s even. For example, If n = 2, then: s(1) = 1, and s(2) = 1, so we have an odd number of sign changes. If n = 3, then: s(1) = 1, s(2) = 1, and s(3) = 1 so we have an even number of sign changes. Statement one alone is not sufficient. Statement Two Alone: n is odd WIthout knowing the sequence, we do not have enough information to determine an answer. Statement two alone is not sufficient to answer the question. Statements One and Two Together: As we’ve shown in statement one, when n is 3, we see that we have an even number of sign changes. When n = 5, the sequence s(k) is: 1, 1, 1, 1, 1, which have 4 sign changes. If we increase n from 5 to 7, we will have 2 more sign changes, therefore, we can see that if n is odd, the sequence s(k) = (1)^k will always produce an even number of sign changes. Answer: C
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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07 May 2019, 22:15
ScottTargetTestPrep wrote: Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION s(k) = (1)^k We see that s(k) is the sequence: 1, 1, 1, 1. ... , (1)^n However, since we don’t know the value of n, we can’t determine the number of sign changes or whether it’s even. For example, If n = 2, then: s(1) = 1, and s(2) = 1, so we have an odd number of sign changes. If n = 3, then: s(1) = 1, s(2) = 1, and s(3) = 1 so we have an even number of sign changes. Statement one alone is not sufficient. Statement Two Alone: n is odd WIthout knowing the sequence, we do not have enough information to determine an answer. Statement two alone is not sufficient to answer the question. Statements One and Two Together: As we’ve shown in statement one, when n is 3, we see that we have an even number of sign changes. When n = 5, the sequence s(k) is: 1, 1, 1, 1, 1, which have 4 sign changes. If we increase n from 5 to 7, we will have 2 more sign changes, therefore, we can see that if n is odd, the sequence s(k) = (1)^k will always produce an even number of sign changes. Answer: C I dont understand why the knowledge sequence is required isnt total no. of term sufficient to calculate answer? pls explain with example if any.



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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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10 May 2019, 17:32
Mohammad Ali Khan wrote: ScottTargetTestPrep wrote: Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION s(k) = (1)^k We see that s(k) is the sequence: 1, 1, 1, 1. ... , (1)^n However, since we don’t know the value of n, we can’t determine the number of sign changes or whether it’s even. For example, If n = 2, then: s(1) = 1, and s(2) = 1, so we have an odd number of sign changes. If n = 3, then: s(1) = 1, s(2) = 1, and s(3) = 1 so we have an even number of sign changes. Statement one alone is not sufficient. Statement Two Alone: n is odd WIthout knowing the sequence, we do not have enough information to determine an answer. Statement two alone is not sufficient to answer the question. Statements One and Two Together: As we’ve shown in statement one, when n is 3, we see that we have an even number of sign changes. When n = 5, the sequence s(k) is: 1, 1, 1, 1, 1, which have 4 sign changes. If we increase n from 5 to 7, we will have 2 more sign changes, therefore, we can see that if n is odd, the sequence s(k) = (1)^k will always produce an even number of sign changes. Answer: C I dont understand why the knowledge sequence is required isnt total no. of term sufficient to calculate answer? pls explain with example if any.The key here is that we don’t know the total number of the terms in the sequence. It just says there are n terms in the sequence, but we don’t know the value of n. Now even if we knew the value of n, we still would not know the number of sign changes unless we knew how the terms were defined. The sequence 2, 3, 4, 5, is just an example. It’s neither the actual sequence nor does it indicate how the actual sequence will behave (i.e., don’t expect the actual sequence to alternate the signs just like the forexample sequence),. To put it another way, if we only know statement one, how would you be able to tell whether the sequence is {1, 1, 1} or {1, 1, 1, 1}? The first one has two sign changes, and the second one has three sign changes. If we only know that n is odd, in other words, if we only know that there is an odd number of terms in the sequence, how can we tell whether the sequence is {3, 5, 4, 2, 7} or {6, 2, 1}? (Remember that we don’t know anything about the sequence, we only know that it has an odd number of terms.) In the first example, there are no sign changes, but in the second example, there is one sign change. It is only when we know both statements that we know for sure that there is an even number of sign changes.
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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11 May 2019, 00:58
Mohammad Ali Khan wrote: ScottTargetTestPrep wrote: Bunuel wrote: In any sequence of n nonzero numbers, a pair of consecutive terms with opposite signs represents a sign change. For example, the sequence –2, 3, –4, 5 has three sign changes. Does the sequence of nonzero numbers s1, s2, s3, . . . , sn have an even number of sign changes?
(1) \(s_k = (–1)^k\) for all positive integers k from 1 to n. (2) n is odd.
DS57502.01 OG2020 NEW QUESTION s(k) = (1)^k We see that s(k) is the sequence: 1, 1, 1, 1. ... , (1)^n However, since we don’t know the value of n, we can’t determine the number of sign changes or whether it’s even. For example, If n = 2, then: s(1) = 1, and s(2) = 1, so we have an odd number of sign changes. If n = 3, then: s(1) = 1, s(2) = 1, and s(3) = 1 so we have an even number of sign changes. Statement one alone is not sufficient. Statement Two Alone: n is odd WIthout knowing the sequence, we do not have enough information to determine an answer. Statement two alone is not sufficient to answer the question. Statements One and Two Together: As we’ve shown in statement one, when n is 3, we see that we have an even number of sign changes. When n = 5, the sequence s(k) is: 1, 1, 1, 1, 1, which have 4 sign changes. If we increase n from 5 to 7, we will have 2 more sign changes, therefore, we can see that if n is odd, the sequence s(k) = (1)^k will always produce an even number of sign changes. Answer: C I dont understand why the knowledge sequence is required isnt total no. of term sufficient to calculate answer? pls explain with example if any. I asked just because the question itself was just concerned with whether seq. has even or odd number of sign changes it never asked anything about sequence and its values.This confused me.



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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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11 May 2019, 14:15
Hi All, We're told that in any sequence of N NONZERO numbers, a pair of consecutive terms with opposite signs represents a sign change (for example, the sequence 2, 3, 4, 5 has three sign changes). We're asked if the sequence of nonzero numbers s1, s2, s3, . . . , sn has an EVEN number of sign changes. This is a YES/NO question. This question can be solved with a bit of Arithmetic and TESTing VALUES. (1) sk = (1)^k for all positive integers k from 1 to N. While the information in Fact 1 might look complex, it actually defines how the sequence 'works.' The initial terms  and the pattern behind the sequence  are... 1st term = (1)^1 = 1 2nd term = (1)^2 = +1 3rd term = (1)^3 = 1 4th term = (1)^4 = +1 Etc. We know how the sequence repeats, but we DON'T know how many terms are in it, so we don't know exactly how many sign changes there are (and thus, we don't know if the total number of changes is even OR odd). IF.... There are 3 terms, then we have 1, +1, 1 and two sign changes  so the answer to the question is YES. There are 4 terms, then we have 1, +1, 1, +1 and three sign changes  so the answer to the question is NO. Fact 1 is INSUFFICIENT (2) N is ODD. Fact 2 tells us the number of terms in the sequence, but we know nothing about any of their values, so we don't know how many sign changes occur. Fact 2 is INSUFFICIENT Combined, we know... sk = (1)^k for all positive integers k from 1 to N. N is ODD. We know exactly how the sequence 'works' and that there are an ODD number of terms. IF There are 3 terms, then we have 1, +1, 1 and two sign changes  so the answer to the question is YES. There are 5 terms, then we have 1, +1, 1, +1, 1 and four sign changes  so the answer to the question is YES. There are 7 terms, then we have 1, +1, 1, +1, 1, +1, 1 and six sign changes  so the answer to the question is YES. Etc. Whenever there are an odd number of terms, the answer to the question is YES. Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: In any sequence of n nonzero numbers, a pair of consecutive terms with
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