In an increasing geometric progression, the difference between the fifth and first terms is 15 and the difference

In an increasing geometric progression, the difference between the fifth and first terms is 15 and the difference between the fourth and second terms is 6. The sum of all members of the progression is 127. Determine the number of members in this progression.

By condition
(1) b5 – b1 = 15;
(2) b4 – b2 = 6.
We use the following formulas:
bn = b1 * q ^ (n-1), n ​​= 1, 2 …
Sn = b1 * (q ^ n – 1) / (q – 1).
Taking into account the formulas, we rewrite equations (1) and (2):
b1 * q ^ 4 – b1 = b1 * (q ^ 4 – 1) = b1 * (q ^ 2 – 1) * (q ^ 2 + 1) = 15;
b1 * q ^ 3 – b1 * q = b1 * q * (q ^ 2 – 1) = 6;
Divide the first by the second and find q.
b1 * (q ^ 2 – 1) * (q ^ 2 + 1) / b1 * q * (q ^ 2 – 1) = 15/6.
(q ^ 2 + 1) / q = 15/6.
6 * q ^ 2 + 6 = 15 * q.
2 * q ^ 2 – 5 * q + 2 = 0.
Let’s find the discriminant.
q1,2 = (5 ± (5 ^ 2 – 4 * 2 * 2) ^ (1/2) / 4 = (5 ± 3) / 4.
q1 = 2.
q2 = ½ does not fit, since in our case the progression is increasing and q> 1.
Substitute the values ​​in the formulas b1 and Sn = 127.
b1 = 15 / (q ^ 4 – 1) = 15 / (2 ^ 4 – 1) = 1.
127 = 1 * (2 ^ n – 1) / (2 -1) = 2 ^ n – 1.
128 = 2 ^ n.
N = 7.
There are 7 members in this progression.