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A series has three numbers a, ar, and ar2. In the series, th
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02 May 2019, 02:45
02 May 2019, 02:45
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A sequence has three numbers \(a\), \(ar\), and \(ar^2\). In the sequence, the first term is twice the second term. What is the ratio of the sum of the first two terms to the sum of the last two terms in the sequence?
(A) 1 : 1
(B) 1 : 2
(C) 1 : 4
(D) 2 : 1
(E) 4 : 1
(A) 1 : 1
(B) 1 : 2
(C) 1 : 4
(D) 2 : 1
(E) 4 : 1
Re: A series has three numbers a, ar, and ar2. In the series, th
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04 May 2019, 07:44
04 May 2019, 07:44
is it like, say, first term 2, second term 1. So, we find a common ratio. therefore, the third term .5 .
So, sum of first two terms =3 while the sum of the last two terms 1.5
thus, the ratio is 3/1.5 or 2/1?
So, sum of first two terms =3 while the sum of the last two terms 1.5
thus, the ratio is 3/1.5 or 2/1?
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Re: A series has three numbers a, ar, and ar2. In the series, th
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04 May 2019, 07:51
04 May 2019, 07:51
1
Carcass wrote:
A sequence has three numbers \(a\), \(ar\), and \(ar^2\). In the sequence, the first term is twice the second term. What is the ratio of the sum of the first two terms to the sum of the last two terms in the sequence?
(A) 1 : 1
(B) 1 : 2
(C) 1 : 4
(D) 2 : 1
(E) 4 : 1
(A) 1 : 1
(B) 1 : 2
(C) 1 : 4
(D) 2 : 1
(E) 4 : 1
The first term is twice the second term
In other words: term1/term2 = 2
Or we can write: \(\frac{a}{ar}=2\)
Simplify: \(\frac{1}{r}=2\)
Solve: \(r=\frac{1}{2}\)
Now that we know \(r=\frac{1}{2}\), the 3 numbers in the sequence are as follows:
term1 \(=a\),
term2 \(=ar=a(\frac{1}{2})=\frac{a}{2}\)
term3 \(=ar^2=a(\frac{1}{2})^2=a(\frac{1}{4})=\frac{a}{4}\)
What is the ratio of the sum of the first two terms to the sum of the last two terms in the sequence?
Sum of the first two terms \(= a + \frac{a}{2}= \frac{2a}{2} + \frac{a}{2}= \frac{3a}{2}\)
Sum of the last two terms \(= \frac{a}{2}+\frac{a}{4}= \frac{2a}{4} + \frac{a}{4}= \frac{3a}{4}\)
ratio = (3a/2)/(3a/4)
= (3a/2)(4/3a)
= 12a/6a
= 2/1
= 2:1
Answer: D
Cheers,
Brent
