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]

06 Nov 2009, 15:44

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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, 05:59, edited 1 time in total.

Renamed the topic.

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Re: Sequence2 [#permalink]

06 Nov 2009, 16:22

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]

07 Nov 2009, 23:53

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]

28 Sep 2012, 05:56

Is this a 700 Category question ??
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Re: Sequence2 [#permalink]

04 Nov 2012, 13:22

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]

04 Nov 2012, 22:17

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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+(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]

05 Nov 2012, 02:54

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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]

02 Dec 2012, 08:56

bhavinshah5685 wrote:

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.

Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find [#permalink] 02 Dec 2012, 08:56

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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

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

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