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If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find

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If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find [#permalink] New post 06 Nov 2009, 14:44
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Sep 2012, 04:59, edited 1 time in total.
Renamed the topic.
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Re: Sequence2 [#permalink] New post 06 Nov 2009, 15:22
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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 = \(n^2\)

\(\frac{50*51}{n^2}=\frac{51}{2}\)

\(n^2=100\)

\(n=10\)
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Re: Sequence2 [#permalink] New post 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 [#permalink] New post 28 Sep 2012, 04:56
Is this a 700 Category question ??
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Re: Sequence2 [#permalink] New post 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 [#permalink] New post 04 Nov 2012, 21:17
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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+(n-1)d}

...

n/2 { 2x1 + (n-1)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 [#permalink] New post 05 Nov 2012, 01:54
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Sn=n/2(2a+(n-1)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 [#permalink] New post 02 Dec 2012, 07:56
bhavinshah5685 wrote:
Sn=n/2(2a+(n-1)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 [#permalink] New post 03 Aug 2014, 11:43
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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find [#permalink] New post 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!
:oops: :oops:
Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find   [#permalink] 09 May 2015, 04:28
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