You are given a geometric progression (bn), the denominator of which is 4 b1 = 3/4 Find the sum of its first 6 terms.

To find the sum of the first six terms of a given geometric progression, we use the formula for the sum of the first n terms of a geometric progression Sn = b1 * (1-q ^ n) / (1-q), where b1 is the first term of a geometric progression, q is the denominator of a geometric progression. According to the condition of the problem, b1 = 3/4, q = 4, therefore for n = 6 we get:
S6 = b1 * (1-q ^ 6) / (1-q) = (3/4) * (1-4 ^ 6) / (1-4) = (3/4) * (1-4 ^ 6 ) / (- 3) = (3/4) * (- 4095) / (- 3) = (3/4) * 4095/3 = 4095/4 = 1023.75

Answer: the sum of the first six members of this geometric progression is 1023.75.