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A series has three numbers, a, ar and ar^2. In the series, the first..
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29 Sep 2017, 04:05
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Question Stats:
68% (01:36) correct 32% (01:36) wrong based on 147 sessions
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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?
Re: A series has three numbers, a, ar and ar^2. In the series, the first..
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29 Sep 2017, 04:49
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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 \(= (a+\frac{a}{2})/(\frac{a}{2}+\frac{a}{4})=\frac{2}{1}\)
Re: A series has three numbers, a, ar and ar^2. In the series, the first..
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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.
Re: A series has three numbers, a, ar and ar^2. In the series, the first..
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15 Nov 2017, 21:21
1
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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A series has three numbers, a, ar and ar^2. In the series, the first..
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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..
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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..
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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..
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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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