The first term of the geometric progression is 3 (2-√2), and the sum of its first two members is 3

The first term of the geometric progression is 3 (2-√2), and the sum of its first two members is 3. Find the sum of all the members of the progression.

1. B1 and S2 are given exponentially bn:

b1 = 3 (2 – √2);
S2 = 3.
2. Find the second term and denominator of the progression:

b1 + b2 = S2;
3 (2 – √2) + b2 = 3;
6 – 3√2 + b2 = 3;
b2 = 3 – 6 + 3√2;
b2 = -3 + 3√2 = 3 (√2 – 1);
q = b2 / b1;
q = 3 (√2 – 1) / 3 (2 – √2) = (√2 – 1) / (2 – √2) = (√2 – 1) (2 + √2) / (4 – 2) = (2√2 + 2 – 2 – √2) / 2 = √2 / 2 = 1 / √2.
3. The sum of all members of the progression is found by the formula:

S = b1 / (1 – q);
S = 3 (2 – √2) / (1 – 1 / √2) = 3 (2 – √2) √2 / (√2 – 1) = 3 * (√2) ^ 2 (√2 – 1) / (√2 – 1) = 6.
Answer: 6.