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#mathematics
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#geometric-progression
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A) The sum of infinity of GP is four times the first term. Find the common ratio?
B) The limiting sum of a GP is twice the sum of the first two terms. Find the possible values of the common ratio?

Mar 24

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Mar 24

A) Let the first term of the GP be a and the common ratio be r.

The sum of an infinite GP is given by S = a / (1 - r)

Given that the sum is four times the first term:

4a = a / (1 - r)

4 = 1 / (1 - r)

4 - 4r = 1

4r = 3

r = 3/4

Therefore, the common ratio is 3/4.

B) Let the first two terms of the GP be a and ar.

The sum of the first two terms is given by S2 = a + ar = a(1 + r)

Given that the limiting sum is twice the sum of the first two terms:

2(a / (1 - r)) = a(1 + r)

2 / (1 - r) = 1 + r

2 = (1 - r)(1 + r)

2 = 1 - r^2

r^2 = -1

Since the common ratio of a geometric progression must be positive, there are no possible values of the common ratio that satisfy this condition.

The sum of an infinite GP is given by S = a / (1 - r)

Given that the sum is four times the first term:

4a = a / (1 - r)

4 = 1 / (1 - r)

4 - 4r = 1

4r = 3

r = 3/4

Therefore, the common ratio is 3/4.

B) Let the first two terms of the GP be a and ar.

The sum of the first two terms is given by S2 = a + ar = a(1 + r)

Given that the limiting sum is twice the sum of the first two terms:

2(a / (1 - r)) = a(1 + r)

2 / (1 - r) = 1 + r

2 = (1 - r)(1 + r)

2 = 1 - r^2

r^2 = -1

Since the common ratio of a geometric progression must be positive, there are no possible values of the common ratio that satisfy this condition.