In a geometric progression, b3 = 12, b6 = -96. Find the sum of the first five terms of this progression.

Find the first term b1 and the denominator q of this geometric progression.

By the condition of the problem, the third term b3 of this geometric sequence is 12, in the sixth term b6 of this geometric sequence is 96.

Using the formula for the nth term of the geometric progression bn = b1 * q ^ (n – 1), we obtain the following relations:

b1 * q ^ (3 – 1) = 12;

b1 * q ^ (6 – 1) = 96.

Dividing the first ratio by the first, we get:

b1 * q ^ 5 / (b1 * q ^ 2) = 96/12;

q ^ 5 / q ^ 2 = 8;

q ^ 3 = 8;

q = 2.

Substituting the found value q = 2 into the equation b1 * q ^ 2 = 12, we get:

b1 * 2 ^ 2 = 12;

b1 * 4 = 12;

b1 = 12/4;

b1 = 3.

Using the formula for the sum of the first n terms of the geometric progression Sn = b1 * (1 – q ^ n) / (1 – q) for n = 5, we find the sum of the first five terms of this progression:

S5 = 3 * (1 – 2 ^ 5) / (1 – 2) = 3 * (1 – 32) / (-1) = 3 * (-31) / (-1) = 3 * 31 = 93.

Answer: The sum of the first five members of this progression is 93.