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A series has three numbers, a, ar and ar^2. In the series, the first..

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A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 29 Sep 2017, 03:05
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A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1

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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 29 Sep 2017, 03:49
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fitzpratik wrote:
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1


Given \(a=2ar\), or \(r=\frac{1}{2}\)

Hence the terms are \(a\), \(\frac{a}{2}\) & \(\frac{a}{4}\)

Ratio \(= (a+\frac{a}{2})/(\frac{a}{2}+\frac{a}{4})=\frac{2}{1}\)

Option D
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 08 Oct 2017, 11:29
fitzpratik wrote:
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1


The ans should be 2:1.
Though I read the question incorrect as the first term being half of the second term.
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 14 Nov 2017, 15:18
Wouldn't ar^2 become a^2/4?
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 14 Nov 2017, 18:44
BigUD94 wrote:
Wouldn't ar^2 become a^2/4?

Hi BigUD94
Here it is r^2 and not a^2, hence it will be a/4 when you substitute the value of r.

Posted from my mobile device
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 15 Nov 2017, 08:06
niks18

Oh alright sir I thought it was (ar)^2 by mistake, I got it now. Thanks
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 15 Nov 2017, 08:24
BigUD94 wrote:
niks18

Oh alright sir I thought it was (ar)^2 by mistake, I got it now. Thanks


Hi BigUD94

you suddenly made me 50 year old guy by writing that "sir" :lol: :lol:
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 15 Nov 2017, 20:21
fitzpratik wrote:
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1


Such Questions can be done in two ways..

Algebraic as shown above
\(2ar=a......r=\frac{1}{2}\)...
So terms are a, a/2 and a/4..
Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1...
2:1

Other slightly easier is substitution, rather smart substitution looking at the question
Let a=4..
So ar will be half, 4/2=2
And ar^2 will become 2/2=1
Number 4,2,1...
We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\)
So 2:1
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 17 Nov 2017, 10:50
niks18

Har koi gyaani hai idhar, mu se sir nikal hi jaata hai out of respect :D
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A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 17 Nov 2017, 16:35
chetan2u wrote:
fitzpratik wrote:
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1


Such Questions can be done in two ways..

Algebraic as shown above
\(2ar=a......r=\frac{1}{2}\)...
So terms are a, a/2 and a/4..
Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1...
2:1

Other slightly easier is substitution, rather smart substitution looking at the question
Let a=4..
So ar will be half, 4/2=2
And ar^2 will become 2/2=1
Number 4,2,1...
We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\)
So 2:1

chetan2u , I like this second method. I'm a little confused by this part:
Quote:
Let a=4..
So ar will be half, 4/2=2
And ar^2 will become 2/2=1

Why is third term derived from 2/2? I thought it would be:
\(a = 4\)
\(a_2 = 4 * \frac{1}{2} = 2\)
\(a_3 = 4 * \frac{1}{4} = 1\)

I'm missing something about what happened to ratio, r, here . . .
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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 17 Nov 2017, 18:37
1
genxer123 wrote:
chetan2u wrote:
fitzpratik wrote:
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?

A. 1:1

B. 1:2

C. 1:4

D. 2:1

E. 4:1


Such Questions can be done in two ways..

Algebraic as shown above
\(2ar=a......r=\frac{1}{2}\)...
So terms are a, a/2 and a/4..
Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1...
2:1

Other slightly easier is substitution, rather smart substitution looking at the question
Let a=4..
So ar will be half, 4/2=2
And ar^2 will become 2/2=1
Number 4,2,1...
We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\)
So 2:1

chetan2u , I like this second method. I'm a little confused by this part:
Quote:
Let a=4..
So ar will be half, 4/2=2
And ar^2 will become 2/2=1

Why is third term derived from 2/2? I thought it would be:
\(a = 4\)
\(a_2 = 4 * \frac{1}{2} = 2\)
\(a_3 = 4 * \frac{1}{4} = 1\)

I'm missing something about what happened to ratio, r, here . . .


Hi..

No you are correct, both mean the same thing...
you have taken \(\frac{4}{4}=4*\frac{1}{4}\) as \(a*r^2\), that is first term * r^2
whereas in my calculations ... \(\frac{2}{2}=2*\frac{1}{2}\) as \(ar*r\), that is 2nd term * r
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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New post 17 Nov 2017, 19:48
chetan2u wrote:
Hi..

No you are correct, both mean the same thing...
you have taken \(\frac{4}{4}=4*\frac{1}{4}\) as \(a*r^2\), that is first term * r^2
whereas in my calculations ... \(\frac{2}{2}=2*\frac{1}{2}\) as \(ar*r\), that is 2nd term * r

chetan2u , thanks! My brain short-circuited. I appreciate the help.
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A series has three numbers, a, ar and ar^2. In the series, the first.. &nbs [#permalink] 17 Nov 2017, 19:48
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