Find the denominator of an infinitely decreasing geometric progression if the sum of all the members

Find the denominator of an infinitely decreasing geometric progression if the sum of all the members of the progression is 2, and the sum of the squares of all the members of this progression is 5.

Let us show that if the sequence bn is a geometric progression with denominator q, then the sequence cn = (bn) ^ 2 is also a geometric progression with denominator q ^ 2.
Let us find the ratio of n + 1 members of the sequence cn to the nth term of this sequence:
cn + 1 / cn = ((bn + 1) ^ 2) / (bn) ^ 2 = ((b1 * q ^ n) ^ 2) / (b1 * q ^ (n-1) ^ 2 = b1 ^ 2 * q ^ 2n / (b1 ^ 2 * q ^ (2n-2)) = q ^ 2n / q ^ (2n-2) = q ^ 2.
Thus, the sequence cn is a geometric progression with the denominator q ^ 2 and the first term c1 = b1 ^ 2.
By the condition of the problem, the sum of all terms of the infinitely decreasing geometric progression bn is equal to 2. Using the formula for the sum of the infinitely decreasing geometric progression S = b1 / (1 – q), we can write:
b1 / (1 – q) = 2.
Since the sequence bn is infinitely decreasing, its denominator q is less than 1. Therefore, the value q ^ 2 is also less than 1, which means that the sequence cn is also infinitely decreasing.
By the condition of the problem, the sum of all members of the progression cn is 5, therefore, we can write:
b1 ^ 2 / (1 – q ^ 2) = 5.
We solve the resulting system of equations.
Dividing the second equation by the first equation, we get:
(b1 ^ 2 / (1 – q ^ 2)) / (b1 / (1 – q)) = 5/2;
(b1 ^ 2 / (1 – q ^ 2)) * ((1 – q) / b1) = 5/2;
b1 ^ 2 * (1 – q) / ((1 – q ^ 2) * b1) = 5/2;
b1 ^ 2 * (1 – q) / ((1 – q) * (1 + q) * b1) = 5/2;
b1 * / (1 + q) = 5/2.
Substituting the value b1 = 2 * (1 – q) into the resulting ratio, we get:
2 * (1 – q) / (1 + q) = 5/2.
We solve the resulting equation:
4 * (1 – q) = 5 * (1 + q);
4 – 4 * q = 5 + 5 * q;
5 * q + 4 * q = 4 – 5;
9 * q = -1;
q = -1/9.

Answer: The denominator of this geometric progression is -1/9.