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The infinite sequence a1, a2,...,an is defined such that an [#permalink]
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19 Feb 2010, 13:03
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The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence? (A) 45 (B) 66 (C) 90 (D) 121 (E) 132
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Re: PS Question 700 level  Need easy to solve this [#permalink]
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19 Feb 2010, 13:09
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nth term = (n+2) / n where n>=1
product of 1st 10 terms ... as you can see, the numeration of the first expression will be same as the denominator of the 3rd expression.
so that leaves us with denominato of the 1 st and 2 nd and numerators of the last and last but one terms
11 * 12 / 1 * 2 = 66
B



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19 Feb 2010, 13:14
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shuj00 wrote: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45 (B) 66 (C) 90 (D) 121 (E) 132 It's very often a good idea in sequence questions to write down the first few terms; typically a pattern will emerge. Here we find each term by plugging n into the expression given: a_1 = (1+2)/1 = 3/1 a_2 = (2 + 2)/2 = 4/2 a_3 = (3 + 2)/3= 5/3 ... a_9 = 11/9 a_10 = 12/10 Now, if we multiply the first 10 terms we have: (3/1) * (4/2) * (5/3) * ... *(10/8)*(11/9)*(12/10) = (3*4*5*...*10*11*12)/(1*2*3*...*8*9*10) Notice that almost the entire denominator can be canceled, leaving us with (11*12)/2 = 66.
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Re: Sum of consecutive terms [#permalink]
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15 Nov 2010, 17:56
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Yalephd wrote: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45 (B) 66 (C) 90 (D) 121 (E) 132
Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem? Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4 Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66
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Re: Sum of consecutive terms [#permalink]
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09 Dec 2010, 08:12
VeritasPrepKarishma wrote: Yalephd wrote: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45 (B) 66 (C) 90 (D) 121 (E) 132
Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem? Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4 Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66 Since this is a sequence and seeing the pattern " increase of numerator and denominator by 1/1" we determine the next term.This part I understood. But fundamental thing that I have not yet grasped is how to visualise that the respective denominators and numerators actually cancel without enumerating them all. I had to list out all the fractions (just by adding 1/1 to the next) to see that they do cancel and what the final fractions would remain (though in matters of time I did not consume much timefew extra seconds). Is there a logic behind this visualisation ? Please advice.



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Re: Sum of consecutive terms [#permalink]
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09 Dec 2010, 11:04
helloanupam wrote: VeritasPrepKarishma wrote: Yalephd wrote: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45 (B) 66 (C) 90 (D) 121 (E) 132
Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem? Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4 Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66 Since this is a sequence and seeing the pattern " increase of numerator and denominator by 1/1" we determine the next term.This part I understood. But fundamental thing that I have not yet grasped is how to visualise that the respective denominators and numerators actually cancel without enumerating them all. I had to list out all the fractions (just by adding 1/1 to the next) to see that they do cancel and what the final fractions would remain (though in matters of time I did not consume much timefew extra seconds). Is there a logic behind this visualisation ? Please advice. The nth term is (n+2)/n. So I say, when I multiply terms, they will look like this: \(\frac{[highlight](n+2)[/highlight]}{n} * \frac{(n+3)}{(n+1)} * \frac{(n+4)}{(n+2)} * \frac{(n+5)}{(n+3)} * ... * \frac{(n+12)}{(n+10)}\) Watch what happens here. The numerator (n+2) of the 1st term has appeared in the denominator of the 3rd term so they will cancel out. So 2nd term will have numerator that is 1 more than the numerator of the 1st term and 4th term will have a denominator that is one more than the denominator of the 3rd term and hence they will be equal and will cancel out. This will continue till we run out of denominators to cancel. So even without enumerating, we know that we will be left with first two numerators and last two denominators. (I wish I could strike the relevant terms with different colored lines but I haven't been able to figure out how to use other operators in fractions!)
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Re: Sum of consecutive terms [#permalink]
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09 Dec 2010, 20:29
How do we know a1=1? we don't know the value of the first 10 terms? how can we assume a1=1? I didn't see that and was confused by the notation.



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Re: Sum of consecutive terms [#permalink]
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09 Dec 2010, 23:56
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gettinit wrote: How do we know a1=1? we don't know the value of the first 10 terms? how can we assume a1=1? I didn't see that and was confused by the notation. \(a_1\) doesn't equal to 1, it equals to \(a_1=\frac{1+2}{1}=\frac{3}{1}\). The infinite sequence a1, a2,..., an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?(A) 45 (B) 66 (C) 90 (D) 121 (E) 132 Stem gives the formula to get the terms of the sequences: \(a_n=\frac{n+2}{n}\) for ALL \(n\geq{1}\) (so starting from \(a_1\)). So: \(a_1=\frac{1+2}{1}=\frac{3}{1}\); \(a_2=\frac{2+2}{2}=\frac{4}{2}\); \(a_3=\frac{3+2}{3}=\frac{5}{3}\); ... \(a_{10}=\frac{10+2}{10}=\frac{12}{10}\); The product of the first 10 terms of the sequence will equal: \(\frac{3}{1}*\frac{4}{2}*\frac{5}{3}*...*\frac{12}{10}=\frac{(3*4*5*6*6*8*9*10)*11*12}{1*2*(3*4*5*6*6*8*9*10)}=\frac{11*12}{2}=66\). Answer: B. Hope it's clear.
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Re: Sum of consecutive terms [#permalink]
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10 Dec 2010, 04:39
Thanks Karishma. I took a couple of bounces before I understood from your explanation albeit in own fashion (and hopefully correct!).The way I understand is this the first two denominators (n & n+1) do not have their mirror numerators and similarly the final two numerators (n+11) & (n+12) wont get to see their mirror denominator images as the process stops at denominator (n+10) and hence dont cancel out.



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Re: Sum of consecutive terms [#permalink]
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10 Dec 2010, 04:52
gettinit wrote: How do we know a1=1? we don't know the value of the first 10 terms? how can we assume a1=1? I didn't see that and was confused by the notation. This is what is given to you in the question: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. The product of first 10 terms is asked. Using the rule, an = (n+2) / n, you can find the nth term easily. The first term will have n = 1 a1 = (1+2)/1 = 3/1 a2 = (2+2)/2 = 4/2 and so on...
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Re: Sum of consecutive terms [#permalink]
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10 Dec 2010, 04:54
helloanupam wrote: Thanks Karishma. I took a couple of bounces before I understood from your explanation albeit in own fashion (and hopefully correct!).The way I understand is this the first two denominators (n & n+1) do not have their mirror numerators and similarly the final two numerators (n+11) & (n+12) wont get to see their mirror denominator images as the process stops at denominator (n+10) and hence dont cancel out. That's absolutely correct.
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The infinite sequence a1, a2,...,an is defined such that an = (n [#permalink]
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26 Dec 2012, 18:49
shuj00 wrote: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45 (B) 66 (C) 90 (D) 121 (E) 132 Seeing that the series is a sequence of fractions. Seeing that the question is asking about the product... we know that factors will cancel out and easily reveal the answer.\(3 * 2 * \frac{5}{3}*\frac{3}{2}*\frac{7}{5}*\frac{8}{6}*\frac{9}{7}*\frac{10}{8}*\frac{11}{9}*\frac{12}{10}\) After carefully cancelling out we would be left with \(11*6 = 66\) Answer: B
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