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How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]
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08 Dec 2011, 09:03
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How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 + ... will form 120 + 121*(3^(1/2))? A. 3 B. 4 C. 5 D. 6 E. 9
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Re: Sequences [#permalink]
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08 Dec 2011, 09:19
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divide series in rational and irrational no's since sum of rational is always rational and irrational is always irrational that means in the above series 120=3+9+27+81....4 terms 121*(3^1/2)=(1+3+9+27+81)*(3^1/2).... 5 terms so ans is 9....E



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Re: Sequences [#permalink]
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08 Dec 2011, 14:46
avenkatesh007 wrote: divide series in rational and irrational no's since sum of rational is always rational and irrational is always irrational That's true in this question, but not in general. The sum of two rational numbers is always rational, but the sum of two irrational numbers does not need to be irrational. For example, √2 and 2  √2 are both irrational, but their sum is 2, a rational number. That's not the kind of thing the GMAT ever tests, however.
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Re: Sequences [#permalink]
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23 Apr 2012, 10:41
I did not understand the above explanation. Can someone please explain?



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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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24 Apr 2012, 12:49
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Anasthaesium wrote: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?
A. 3 B. 4 C. 5 D. 6 E. 9 We have a following sequence: \(\sqrt{3}\), \(3\), \(3\sqrt{3}\), \(9\), ... Notice that this sequence is a geometric progression with first term equal to \(\sqrt{3}\) and common ratio also equal to \(\sqrt{3}\). The question asks: the sum of how many terms of this sequence adds up to \(120+121\sqrt{3}\). The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{\sqrt{3}(\sqrt{3}^n1)}{\sqrt{3}1}=120+121\sqrt{3}\) > \(\sqrt{3}(\sqrt{3}^n1)=120\sqrt{3}+121*3120121\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n\sqrt{3}=243\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n=243\) > \(3^{\frac{1+n}{2}}=3^5\) > \(\frac{1+n}{2}=5\) > \(n=9\). Answer: E. Shortcut solution:Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: \(\frac{3(3^k1)}{31}=120\) > \(3^k1=80\) > \(3^k=81\) > \(k=4\), so the number of terms must be either \(2k=8\) (if there are equal number of irrational and rational terms) or \(2k+1=9\) (if # of irrational terms, with \(\sqrt{3}\), is one more than # rational terms), since only 9 is present among options then it must be a correct answers. Hope it's clear.
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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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Anasthaesium wrote: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?
A. 3 B. 4 C. 5 D. 6 E. 9 In a series question, it is always a good idea to write the first few terms to figure out what the series looks like: r = \(\sqrt{3} + 3 + 3\sqrt{3} + 9 + 9\sqrt{3} + 27 + 27\sqrt{3} + ....\) Sum of first 2 terms\(= 3 + \sqrt{3}\) Sum of first 3 terms \(= 3 + (\sqrt{3} + 3\sqrt{3})\) Sum of first 4 terms\(= (3+9) + 4\sqrt{3}\) Sum of first 5 terms \(= 12 + (4 + 9)\sqrt{3}\) etc You want a sum of \(120 + 121\sqrt{3}\) You will obtain 120 by adding 3 + 9 + 27 + 81 (4 terms) and \(121\sqrt{3}\) by adding \((1 + 3 + 9 + 27 + 81)\sqrt{3}\) (i.e. 5 terms) So you need to add a total of 9 terms.
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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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05 Jun 2013, 06:26
My method is not very sophisticated and most probably not the best but in some cases it works. The most crucial part here is to understand the question. Basically we have a sequesnce \sqrt{3}+3+3\sqrt{3}+ .... which is the same as: \sqrt{3}+\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}*\sqrt{3}+.... As it could be seen each time next term is multiplied by \sqrt{3}. Now the questions asks us to find how many numbers this sequence should consists of in order to be equal of 120 + 121*\sqrt{3}. What i did, i have just tried to make the calculation easier and tried to simplify the 120 + 121*\sqrt{3}=40*\sqrt{3}*\sqrt{3}+121*\sqrt{3}=\sqrt{3}(40\sqrt{3}+121) Now go back to our sequence, if we take the common \sqrt{3} out of the brackets we have \sqrt{3}(1+\sqrt{3}+3+3\sqrt{3}+9+9\sqrt{3}) > \sqrt{3}(13+13\sqrt{3}) This is the sum of 6 terms in the sequence, which is clearly less than \sqrt{3}(40\sqrt{3}+121). We can go and check whether 9 (next possible choice) will fit, but considering that 9 is the only choice left it should be it. The answer is E.
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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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05 Jun 2013, 07:42
ziko wrote: My method is not very sophisticated and most probably not the best but in some cases it works. The most crucial part here is to understand the question. Basically we have a sequesnce \sqrt{3}+3+3\sqrt{3}+ .... which is the same as: \sqrt{3}+\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}*\sqrt{3}+....
As it could be seen each time next term is multiplied by \sqrt{3}.
Now the questions asks us to find how many numbers this sequence should consists of in order to be equal of 120 + 121*\sqrt{3}.
What i did, i have just tried to make the calculation easier and tried to simplify the 120 + 121*\sqrt{3}=40*\sqrt{3}*\sqrt{3}+121*\sqrt{3}=\sqrt{3}(40\sqrt{3}+121)
Now go back to our sequence, if we take the common \sqrt{3} out of the brackets we have \sqrt{3}(1+\sqrt{3}+3+3\sqrt{3}+9+9\sqrt{3}) > \sqrt{3}(13+13\sqrt{3}) This is the sum of 6 terms in the sequence, which is clearly less than \sqrt{3}(40\sqrt{3}+121). We can go and check whether 9 (next possible choice) will fit, but considering that 9 is the only choice left it should be it. The answer is E. Please read this post: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (ptess m button for formulas).
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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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11 Mar 2014, 21:31
Is the original question missing a "+9" in the portion describing the sequence? It was hard to know what the series would with just 3 terms (as the question currently is). For example, if we only know 3^(1/2), 3, 3*3^(1/2), one could interpret that the series is defined as every input starting from the 3rd is the product of the previous 2. Hope this makes sense. Please edit the question stem because I see a +9 in the answer explanations but not the question itself.



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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]
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11 Mar 2014, 22:18



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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]
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03 Jul 2014, 12:01
Bunuel wrote: Anasthaesium wrote: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?
A. 3 B. 4 C. 5 D. 6 E. 9 We have a following sequence: \(\sqrt{3}\), \(3\), \(3\sqrt{3}\), \(9\), ... Notice that this sequence is a geometric progression with first term equal to \(\sqrt{3}\) and common ratio also equal to \(\sqrt{3}\). The question asks: the sum of how many terms of this sequence adds up to \(120+121\sqrt{3}\). The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{\sqrt{3}(\sqrt{3}^n1)}{\sqrt{3}1}=120+121\sqrt{3}\) > \(\sqrt{3}(\sqrt{3}^n1)=120\sqrt{3}+121*3120121\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n\sqrt{3}=243\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n=243\) > \(3^{\frac{1+n}{2}}=3^5\) > \(\frac{1+n}{2}=5\) > \(n=9\). Answer: E. Shortcut solution:Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: \(\frac{3(3^k1)}{31}=120\) > \(3^k1=80\) > \(3^k=81\) > \(k=4\), so the number of terms must be either \(2k=8\) (if there are equal number of irrational and rational terms) or \(2k+1=9\) (if # of irrational terms, with \(\sqrt{3}\), is one more than # rational terms), since only 9 is present among options then it must be a correct answers. Hope it's clear. I got confused on how you got from \(\frac{\sqrt{3}(\sqrt{3}^n1)}{\sqrt{3}1}=120+121\sqrt{3}\) > \(\sqrt{3}(\sqrt{3}^n1)=120\sqrt{3}+121*3120121\sqrt{3}\) to > \(\sqrt{3}*\sqrt{3}^n\sqrt{3}=243\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n=243\) > \(3^{\frac{1+n}{2}}=3^5\) > \(\frac{1+n}{2}=5\) > \(n=9\). So the part after the cross multiplication, can you break it down further for me? Like what happened to the 121 *3 ?



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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]
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03 Jul 2014, 12:31
sagnik2422 wrote: Bunuel wrote: Anasthaesium wrote: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?
A. 3 B. 4 C. 5 D. 6 E. 9 We have a following sequence: \(\sqrt{3}\), \(3\), \(3\sqrt{3}\), \(9\), ... Notice that this sequence is a geometric progression with first term equal to \(\sqrt{3}\) and common ratio also equal to \(\sqrt{3}\). The question asks: the sum of how many terms of this sequence adds up to \(120+121\sqrt{3}\). The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{\sqrt{3}(\sqrt{3}^n1)}{\sqrt{3}1}=120+121\sqrt{3}\) > \(\sqrt{3}(\sqrt{3}^n1)=120\sqrt{3}+121*3120121\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n\sqrt{3}=243\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n=243\) > \(3^{\frac{1+n}{2}}=3^5\) > \(\frac{1+n}{2}=5\) > \(n=9\). Answer: E. Shortcut solution:Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: \(\frac{3(3^k1)}{31}=120\) > \(3^k1=80\) > \(3^k=81\) > \(k=4\), so the number of terms must be either \(2k=8\) (if there are equal number of irrational and rational terms) or \(2k+1=9\) (if # of irrational terms, with \(\sqrt{3}\), is one more than # rational terms), since only 9 is present among options then it must be a correct answers. Hope it's clear. I got confused on how you got from \(\frac{\sqrt{3}(\sqrt{3}^n1)}{\sqrt{3}1}=120+121\sqrt{3}\) > \(\sqrt{3}(\sqrt{3}^n1)=120\sqrt{3}+121*3120121\sqrt{3}\) to > \(\sqrt{3}*\sqrt{3}^n\sqrt{3}=243\sqrt{3}\) > \(\sqrt{3}*\sqrt{3}^n=243\) > \(3^{\frac{1+n}{2}}=3^5\) > \(\frac{1+n}{2}=5\) > \(n=9\). So the part after the cross multiplication, can you break it down further for me? Like what happened to the 121 *3 ? \(120\sqrt{3}+121*3120121\sqrt{3}\); \((121*3120)+(120\sqrt{3}121\sqrt{3})\); \(243\sqrt{3}\). \(\sqrt{3}*\sqrt{3}^n=243\); \(3^{\frac{1}{2}}*3^{\frac{n}{2}}=3^5\); \(3^{\frac{1+n}{2}}=3^5\); \(\frac{1+n}{2}=5\); \(n=9\). Hope it's clear now.
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How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]
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10 Sep 2014, 02:10
Anasthaesium wrote: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 + ... will form 120 + 121*(3^(1/2))?
A. 3 B. 4 C. 5 D. 6 E. 9 Since four of the of the terms are provided and is = 12 +4*3^(1/2) approx 18, A & B eliminated add 9*3^(1/2) approx 15 so sum approx 33 eliminated once more and D eliminated , leaving E as the correct answer



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Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]
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17 Jan 2017, 20:29
For me, as soon as I see that 1+3+27+81 = 120, x1 +3
I know the the series is either of the form: ? + 3 + ? + 9 + ? + 27 + ? + 81 + ? (9 terms) or ? + 3 + ? + 9 + ? + 27 + ? + 81 (8 terms)...



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