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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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29 Sep 2017, 04:05
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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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29 Sep 2017, 04:49
1
2
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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08 Oct 2017, 12:29
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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14 Nov 2017, 16:18
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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14 Nov 2017, 19:44
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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15 Nov 2017, 09:06
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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15 Nov 2017, 09:24
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"
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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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15 Nov 2017, 21:21
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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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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17 Nov 2017, 11:50
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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17 Nov 2017, 17:35
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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17 Nov 2017, 19:37
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
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A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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17 Nov 2017, 20:48
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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Re: A series has three numbers, a, ar and ar^2. In the series, the first..  [#permalink]

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31 Jan 2019, 20:34
Good night!

Could someone please explain to me how do everyone is arriving to r = 1/2?

Why can't we consider ar as a whole term?

A1 = a = 4
A2 = ar = 2
A3 = ar2 = 4

I mean, do I have to separate different letters in all the sequence questions?

Kind regards!!!
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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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31 Jan 2019, 20:54
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 =2 ar, then r =1/2

Now put back value of r in a, ar ,$$ar^2$$

a + ar = 3a / 2 --------------------(x)
ar + ar^2 = 3a/4 -----------------(y)

Ratio of (x)/(y)

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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31 Jan 2019, 21:01
1
jfranciscocuencag wrote:
Good night!

Could someone please explain to me how do everyone is arriving to r = 1/2?

Why can't we consider ar as a whole term?

A1 = a = 4
A2 = ar = 2
A3 = ar2 = 4

I mean, do I have to separate different letters in all the sequence questions?

Kind regards!!!

Hi jfranciscocuencag

I would like to share my thoughts on this, Its written in the question, that it is a series.

So a ar ar^2, signifies that it is a Geometric Progression, in which the common ratio(r) is equal

ar/a = ar^2/ar --------(x)

So now if you now that relationship, you can even calculate it like that

Its given a= 2 ar, when you substitute this in (x) you can get LHS as 1/2, which will be equal to RHS.

But you need not write all of this, while solving this question, since you know this you can directly solve.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
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A series has three numbers, a, ar and ar^2. In the series, the first..   [#permalink] 31 Jan 2019, 21:01
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