In a geometric progression, the sum of the first and second terms is 60, and the sum of the second

In a geometric progression, the sum of the first and second terms is 60, and the sum of the second and third is 180. Find the sum of the first five members of the progression.

Let b1 be the first term of a given geometric progression, and q be the denominator of this progression.

According to the condition of the problem, the sum of the first and second terms of this given is 60, and the sum of the second and third terms is 180, therefore, we can write the following ratios:

b1 + b1 * q = 60;

b1 * q + b1 * q ^ 2 = 180.

We solve the resulting system of equations.

Dividing the second equation by the first, we get:

(b1 * q + b1 * q ^ 2) / (b1 + b1 * q) = 180/60;

q * (b1 + b1 * q) / (b1 + b1 * q) = 3;

q = 3.

Substituting the found value q = 3 into the equation b1 + b1 * q = 60, we get:

b1 + b1 * 3 = 60;

b1 * 4 = 60;

b1 = 60/4;

b1 = 15.

Knowing d1 and q, we find the sum of the first five terms of the progression:

S5 = 15 * (1 – 4 ^ 5) / (1 – 4) = 15 * (-1023) / (-3) = 15 * 1023/3 = 5 * 1023 = 5115.

Answer: the sum of the first five members of the progression is 5115.