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Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find [#permalink]
04 Nov 2012, 21:17

2

This post received KUDOS

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]
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.

Re: If (2+4+6+....50 terms)/(1+3+5+....n terms)=51/2, then find [#permalink]
03 Aug 2014, 11:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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