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marishachu [46]
3 years ago
5

Consider a geometric sequence with a first term of 4 and a fourth term of -2.916.

Mathematics
1 answer:
Orlov [11]3 years ago
4 0

Answer:

a) Find the common ratio of this sequence.

Answer: -0.82

b) Find the sum to infinity of this sequence.

Answer: 2.2

Step-by-step explanation:

nth term in geometric series is given by 4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{0.549} \\r = 0.82

where

a is the first term

r is the common ratio and

n is the nth term

_________________________________

given

a = 4

4th term = -2.196

let

common ratio of this sequence. be r

4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{-0.549} \\r = -0.82

a) Find the common ratio of this sequence.

answer: -0.82

sum of infinity of geometric sequence is given by = a/(1-r)

thus,

sum to infinity of this sequence = 4/(1-(-0.82) = 4/1.82 = 2.2

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Answer:

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Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

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d = \sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

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Square both sides:

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Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

13a² - 156a - 1404 = 0

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