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The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is :
- a)4/9
- b)2/9
- c)2/3
- d)1/3
Correct answer is option 'C'. Can you explain this answer?
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The sum of an infinite geometric series with positive terms is 3 and t...
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Solution:
Given, the sum of the infinite geometric series with positive terms is 3.
Let a be the first term and r be the common ratio.
The sum of the infinite geometric series is given by S = a/(1 - r)
Given, the sum of the cubes of the terms is 27/19.
The sum of the cubes of the terms of a geometric series is given by S3 = a^3/(1 - r^3)
We know that S = 3 and S3 = 27/19.
Substituting the values in the above equations, we get
3 = a/(1 - r) ...(1)
27/19 = a^3/(1 - r^3) ...(2)
Dividing equation (2) by equation (1), we get
9/19 = a^2/(1 + r + r^2) ...(3)
We know that a and r are positive.
Hence, a^2 < a^2="" +="" ar="" +="" r^2="(a/(1" -="" r))^2="" />
a^2 < />
a^2/(1 + r + r^2) < 9/(r^2(1="" +="" r="" +="" />
Substituting this in equation (3), we get
9/19 < 9/(r^2(1="" +="" r="" +="" />
1/19 < 1/(r^2(1="" +="" r="" +="" />
r^2(1 + r + r^2) < />
r^3 < />
r < />
Let x = r^(1/3)
We get x^3 < />
x^4 < />
x < />
r < (19)^(1/4)^3="" />
Hence, the common ratio of the geometric series is less than 2^(3/2)/3.
Among the given options, only option (c) 2/3 satisfies this condition.
Therefore, the correct answer is option (c) 2/3.
Given, the sum of the infinite geometric series with positive terms is 3.
Let a be the first term and r be the common ratio.
The sum of the infinite geometric series is given by S = a/(1 - r)
Given, the sum of the cubes of the terms is 27/19.
The sum of the cubes of the terms of a geometric series is given by S3 = a^3/(1 - r^3)
We know that S = 3 and S3 = 27/19.
Substituting the values in the above equations, we get
3 = a/(1 - r) ...(1)
27/19 = a^3/(1 - r^3) ...(2)
Dividing equation (2) by equation (1), we get
9/19 = a^2/(1 + r + r^2) ...(3)
We know that a and r are positive.
Hence, a^2 < a^2="" +="" ar="" +="" r^2="(a/(1" -="" r))^2="" />
a^2 < />
a^2/(1 + r + r^2) < 9/(r^2(1="" +="" r="" +="" />
Substituting this in equation (3), we get
9/19 < 9/(r^2(1="" +="" r="" +="" />
1/19 < 1/(r^2(1="" +="" r="" +="" />
r^2(1 + r + r^2) < />
r^3 < />
r < />
Let x = r^(1/3)
We get x^3 < />
x^4 < />
x < />
r < (19)^(1/4)^3="" />
Hence, the common ratio of the geometric series is less than 2^(3/2)/3.
Among the given options, only option (c) 2/3 satisfies this condition.
Therefore, the correct answer is option (c) 2/3.
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The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes ofits terms is 27/19.Then the common ratio of thisseries is :a)4/9b)2/9c)2/3d)1/3Correct answer is option 'C'. Can you explain this answer?
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