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The first term of a GP is A and the common ratio is R

Given that A=2r and the sum to infinity is 4.calculate the third term

To find the third term of the geometric progression (GP), we need to use the formula for the sum of an infinite GP.

The formula for the sum of an infinite GP is:
S = A / (1 - R), where S represents the sum, A represents the first term, and R represents the common ratio.

Given that the sum to infinity is 4, we can set up the equation:
4 = A / (1 - R)

We are also given that A = 2R, so we can substitute this value into the equation:
4 = 2R / (1 - R)

To solve for R, we can multiply both sides of the equation by (1 - R):
4(1 - R) = 2R

Expanding the equation:
4 - 4R = 2R

Combining like terms:
4 = 6R

Dividing both sides of the equation by 6:
R = 4/6 = 2/3

Now that we have found the value of R, we can find the value of A. Since A = 2R, we substitute the value of R into the equation:
A = 2(2/3) = 4/3

To find the third term of the GP, we will multiply the first term (A) by the common ratio (R) squared since it is a geometric progression:
Third term = A * R^2 = (4/3) * (2/3)^2
Third term = (4/3) * (4/9)
Third term = (16/27)

Therefore, the third term of the geometric progression is 16/27.

To find the third term of a geometric progression (GP), we need the first term (A) and the common ratio (R).

In this case, the first term (A) is given as 2R.

We also know that the sum to infinity (Sāˆž) of the GP is 4.

The formula for the sum to infinity of a GP is Sāˆž = A / (1 - R).

Substituting the given values, we have:
4 = 2R / (1 - R)

To solve for R, we multiply both sides of the equation by (1 - R):
4(1 - R) = 2R

Expanding and simplifying:
4 - 4R = 2R
6R = 4
R = 4/6
R = 2/3

Now that we have found the value of R, we can find the first term A:
A = 2R = 2 * 2/3 = 4/3

Finally, to find the third term (T3), we can use the formula for the nth term of a GP, which is Tn = A * R^(n-1). In this case, n = 3:
T3 = A * R^(3-1) = (4/3) * (2/3)^(3-1) = (4/3) * (2/3)^2 = (4/3) * (4/9) = 16/27

Therefore, the third term of the GP is 16/27.

S = a/(1-r)

a = 2r, so

2r/(1-r) = 4
Now just solve for r, and you can then get

A3 = ar^2