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

Since the sequence bn is a geometric progression, the following relationship is true: bn + 1 = bn * q, where q is the denominator of the geometric progression, bn + 1 is the n + 1-th member of this progression, bn is the n-th member of this progression.
Using this ratio, we find the first six terms of this geometric progression, and then sum them up. By the condition of the problem, q = 4, b1 = 3/4, therefore, we can write:
b1 = 3/4;
b2 = b1 * q = (3/4) * 4 = 3;
b3 = b2 * q = 3 * 4 = 12;
b4 = b3 * q = 12 * 4 = 48;
b5 = b4 * q = 48 * 4 = 192;
b6 = b5 * q = 192 * 4 = 768.
We find the sum of the first six members of this progression:
S6 = b1 + b2 + b3 + b4 + b5 + b6 = 3/4 + 3 + 12 + 48 + 192 + 768 = 1023.75.

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