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If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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Updated on: 28 Sep 2012, 04:59
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70% (01:57) correct 30% (02:22) wrong based on 208 sessions
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If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find value of n A. 12 B. 13 C. 9 D. 10
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Originally posted by virtualanimosity on 06 Nov 2009, 14:44.
Last edited by Bunuel on 28 Sep 2012, 04:59, edited 1 time in total.
Renamed the topic.




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Re: Sequence2
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06 Nov 2009, 15:22




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Re: Sequence2
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07 Nov 2009, 22:53
virtualanimosity wrote: Q.If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find value of n
1.12 2.13 3.9 4.10 Sum of 2+4+....+100 = 2(1+2+3+....+50) = 2(50x51/2) = 50x51 (50x51)/(1+3+5+....n terms) = 51/2 100 = 1+3+5+....n terms 100 = n (First term + last term)/2 100 = n (1 + 19)/2 n = 10
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Re: Sequence2
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28 Sep 2012, 04:56
Is this a 700 Category question ??
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Re: Sequence2
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04 Nov 2012, 12:22
Bunuel wrote: virtualanimosity wrote: Q.If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find value of n
1.12 2.13 3.9 4.10 Nominator sum of first n even integers = \(n*(n+1)=50*51\) Denominator sum of first n odd integers = \(n^2\) \(\frac{50*51}{n^2}=\frac{51}{2}\) \(n^2=100\) \(n=10\) Hi Bunuel, How did you know that the sum of the terms in the denominator was \(n^2\)?



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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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04 Nov 2012, 21:17
Quote: How did you know that the sum of the terms in the denominator was ?
Sum of n terms of an A.P. = n/2 { 2a+(n1)d} ... n/2 { 2x1 + (n1)2} : n/2 { 2 + 2n  2} : n/2 x 2n = n^2 Because the Sum of the first 50 even numbers can be calculated using the same formula to be 2550 , we get the equation: 2550/n^2 = 51/2 and solving for n we get n = 10. You could also attack this question by plugging in the values in the answer choices : Given one choice is 10 , we can easily write down the first 10 odd numbers to be : 1 3 5 7 9 11 13 15 17 19 .. So their sum = 10/5 x (A+L) = 10/5 x 20 = 40 ... Divide 2550 by 40 to get the desired ratio... Hope this helps
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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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05 Nov 2012, 01:54
Sn=n/2(2a+(n1)d)...(1)
Here (2+4+...+50th Term) n=50 a=2 d=2 Putting this in eqn (1) we get numerator = 25*102
Now for denominator, n=n a=1 d=2 putting this values in eqn(1) we get denominator = n*n
Now keeping in question stem, 25*102/n*n = 51/2
n = 10



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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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02 Dec 2012, 07:56
bhavinshah5685 wrote: Sn=n/2(2a+(n1)d)...(1)
Here (2+4+...+50th Term) n=50 a=2 d=2 Putting this in eqn (1) we get numerator = 25*102
Now for denominator, n=n a=1 d=2 putting this values in eqn(1) we get denominator = n*n
Now keeping in question stem, 25*102/n*n = 51/2
n = 10 I can't tell based on this equation what a is in the highlighted equation and what operation we are performing on it. Based on what seems to be happening, it is the first number of the sequence and that is it, the 2 didn't appear to do anything in this equation. help appreciated.



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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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09 May 2015, 04:28
GMAT TIGER wrote: virtualanimosity wrote: Q.If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find value of n
1.12 2.13 3.9 4.10 Sum of 2+4+....+100 = 2(1+2+3+....+50) = 2(50x51/2) = 50x51 (50x51)/(1+3+5+....n terms) = 51/2 100 = 1+3+5+....n terms 100 = n (First term + last term)/2 100 = n (1 + 19)/2 n = 10 Hey , how did u get 19 as last term? please explain!



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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find
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19 Oct 2018, 09:21
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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find &nbs
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