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In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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07 Apr 2015, 06:27
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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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07 Apr 2015, 06:56
Bunuel wrote: In the sequence a1, a2, a3, an, an is determined for all values n > 2 by taking the average of all terms a1 through an1. If a1 = 1 and a3 = 5, then what is the value of a20?
(A) 1 (B) 4.5 (C) 5 (D) 6 (E) 9
Kudos for a correct solution. a3 = (a1 + a2) / 2 5 = (1 + x) / 2; x = 9; a2 = 9 a4 = (a1 + a2 + a3) / 3 a4 = (1 + 9 + 5) / 3 = 5 so after a3 we begin to add fives in set and our average soesn't change so any number after a2 will be equal to 5 Answer is C
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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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07 Apr 2015, 09:24
Harley1980 wrote: Bunuel wrote: In the sequence a1, a2, a3, an, an is determined for all values n > 2 by taking the average of all terms a1 through an1. If a1 = 1 and a3 = 5, then what is the value of a20?
(A) 1 (B) 4.5 (C) 5 (D) 6 (E) 9
Kudos for a correct solution. a3 = (a1 + a2) / 2 5 = (1 + x) / 2; x = 9; a2 = 9 a4 = (a1 + a2 + a3) / 3 a4 = (1 + 9 + 5) / 3 = 5 so after a3 we begin to add fives in set and our average soesn't change so any number after a2 will be equal to 5 Answer is C ditto, not much else to add to a pretty straightforward solution.



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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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08 Apr 2015, 03:21
Zhenek wrote: Harley1980 wrote: Bunuel wrote: In the sequence a1, a2, a3, an, an is determined for all values n > 2 by taking the average of all terms a1 through an1. If a1 = 1 and a3 = 5, then what is the value of a20?
(A) 1 (B) 4.5 (C) 5 (D) 6 (E) 9
Kudos for a correct solution. a3 = (a1 + a2) / 2 5 = (1 + x) / 2; x = 9; a2 = 9 a4 = (a1 + a2 + a3) / 3 a4 = (1 + 9 + 5) / 3 = 5 so after a3 we begin to add fives in set and our average soesn't change so any number after a2 will be equal to 5 Answer is C ditto, not much else to add to a pretty straightforward solution. True. I followed the same route to the answer: \(a_3\) = \(\frac{ a_1 + a _2}{2}\) => \(a_2\) = 9 Similarly, \(a_4\) = \(\frac{1+9+5}{3}\) = 5 \(a_5\) = \(\frac{1+9+5+5}{4}\) = 5 and so on. So, essentially, \(a_n\) = 5 for all n >2. However, just to be sure, notice that the series can now be rewritten as: \(a_n\) = \(\frac{1 + 9 + 5(n3)}{n1}\) for all n>2 => \(a_{20}\) = \(\frac{10 + 5*17}{19}\) = 5. So, answer is (C). Is this a valid way to verify the result?



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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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08 Apr 2015, 23:38
Hi trishulpani, Yes, your way is a valid (and great) way to prove that the pattern exists. You're going to come to find that patterns (of some kind) tend to exist in almost all the Quant questions that you'll see on Test Day. As such, part of your training should focus on the ability to spot (and take advantage of) patterns. Sometimes the pattern(s) won't be obvious though, so you have to do the necessary work to 'discover' the pattern. In that way, you should NOT be staring at the screen hoping that the pattern will "come to you" (since you don't have time for that)  instead, get the work on the pad and physically figure out what's there. You'll find that a far more proactive (and effective) way to deal with the ENTIRE GMAT. GMAT assassins aren't born, they're made, Rich
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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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09 Apr 2015, 00:19
EMPOWERgmatRichC wrote: Hi trishulpani,
Yes, your way is a valid (and great) way to prove that the pattern exists. You're going to come to find that patterns (of some kind) tend to exist in almost all the Quant questions that you'll see on Test Day. As such, part of your training should focus on the ability to spot (and take advantage of) patterns. Sometimes the pattern(s) won't be obvious though, so you have to do the necessary work to 'discover' the pattern. In that way, you should NOT be staring at the screen hoping that the pattern will "come to you" (since you don't have time for that)  instead, get the work on the pad and physically figure out what's there. You'll find that a far more proactive (and effective) way to deal with the ENTIRE GMAT.
GMAT assassins aren't born, they're made, Rich That's some great advice. Thanks Rich !



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Re: In the sequence a1, a2, a3, an, an is determined for all values n > 2 [#permalink]
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13 Apr 2015, 07:12
Bunuel wrote: In the sequence a1, a2, a3, an, an is determined for all values n > 2 by taking the average of all terms a1 through \(a_{n1}\). If a1 = 1 and a3 = 5, then what is the value of a20?
(A) 1 (B) 4.5 (C) 5 (D) 6 (E) 9
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:This question is designed to make you waste time trying to decipher it. A certain pattern is established for this sequence, and then the twentieth term is being asked of us. If the sequence has a pattern for all numbers greater than two, and it gave you the first two numbers, then you could deduce the subsequent terms to infinity (and beyond!). However, only the first and third terms are given, so there is at least an extra element of determining the value of the second term. After that, we may need to calculate 16 intermittent items before getting to the 20th value, so it seems like it might be a time consuming affair. As is often the case on the GMAT, once we get going this may be easier than it initially appears. If a1 is 1 and a3 is 5, we actually have enough information to solve a2. The third term of the sequence is defined as the average of the first two terms, thus a3 = (a1 + a2) / 2. This one equation has three variables, but two of them are given in the premise of the question, leading to 5 = (1 + a2) /2. Multiplying both sides by 2, we get 10 = 1 + a2, and thus a2 has to be 9. The first three terms of this sequence are therefore {1, 9, 5}. Now that we have the first three terms and the general case, we should be able to solve a4, a5 and beyond until the requisite a20. The fourth term, a4 is defined as the average of the first three terms. Since the first three terms are {1, 9, 5}, the fourth term will be a4 = (1 + 9 + 5) / 3. This gives us 15/3, which simplifies to 5. A4 is thus equal to 5. Let’s now solve for a5. The same equation must hold for all an, so a5 = (1 + 9 + 5 + 5) /4, which is 20/4, or again, 5. The third, fourth and fifth terms of this sequence are all 5. Perhaps we can decode a pattern without having to calculate the next fourteen numbers (hint: yes you can!). A3 is 5 because that is the average of 1 and 9. Once we found a3, we set off to find subsequent elements, but all of these elements will follow the same pattern. We take the elements 1 and 9, and then find the average of these two numbers, and then average out all three terms. Since a3 was already the average of a1 and a2, adding it to the equation and finding the average will change nothing. A4 will similarly be 5, and adding it into the equation and taking the average will again change nothing. Indeed all of the terms from A3 to A∞ will be equal to exactly 5, and they will have no effect on the average of the sequence. You may have noticed this pattern earlier than element a5, but it can nonetheless be beneficial to find a few concrete terms in order to cement your hypothesis. You can stop whenever you feel comfortable that you’ve cracked the code (there are no style points for calculating all twenty elements). Indeed, it doesn’t matter how many terms you actually calculate before you discover the pattern. The important part is that you look through the answer choices and understand that term a20, like any other term bigger than a3, must necessarily be 5, answer choice C. While understanding the exact relationship between each term on test day is not necessary, it’s important to try and see a few pattern questions during your test prep and understand the concepts being applied. You may not be able to recognize all the common GMAT traps, but if you recognize a few you can save yourself valuable time on questions. If you find yourself faced with a confusing or convoluted question, remember that you don’t have to tackle the problem in a linear fashion. If you’re stuck, try to establish what the key items are, or determine the end and go backwards. When in doubt, don’t be afraid to skip around (figuratively, literal skipping is frowned upon at the test center).
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