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?

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.