Find the denominator of an infinite geometric progression if b1 = 1 and S = 2/3.

In the task, it is required to calculate the denominator (which we denote by q) of an infinite geometric progression, if b1 = 1 and S = 2/3. So, an infinite geometric progression b1, b2, b3, … is given, the first term of which is 1, and the sum is 2 / 3. It is required to determine the value of the denominator q of a given geometric progression.
As you know, an infinitely decreasing geometric progression is a progression in which | q | <1. For it, the concept of the sum of the members of an infinitely decreasing geometric progression is defined as a number to which the sum of the first n terms of the considered progression approaches without bound as the number n increases indefinitely. Moreover, if the denominator q of the geometric progression b1, b2, b3,… satisfies the inequality | q | <1, then the sum of the progression S exists and is calculated by the formula S = b1 / (1 – q).
Since b1 = 1> S = 2/3, it is obvious that q <0. Let’s use the above formula for calculating the sum of an infinitely decreasing geometric progression. We have: 2/3 = 1 / (1 – q). Let’s solve the resulting equation for the unknown q. We have: 1 – q = 1: (2/3), whence, q = 1 – 3/2 = -0.5.
Answer: -0.5.