In a geometric progression, the sum of the first and second terms is 140

In a geometric progression, the sum of the first and second terms is 140, and the sum of the second and third terms is 105. Find these numbers.

Given: b1 + b2 = 140 and b2 + b3 = 105 (bn is the nth term of the geometric progression).
By definition: b2 = b1 * q, b3 = b2 * q,…, bn = bn – 1 * q, where q is the denominator of the progression.
From b2 + b3 = 105 we get b1 * q + b2 * q = 105 or (b1 + b2) * q = 105.
Considering b1 + b2 = 140 from the last equality of item 4: 140 * q = 105, whence q = 105/140 = ¾.
From b1 + b2 = 140: b1 + b1 * q = 140 or b1 * (1 + ¾) = 140. Hence, b1 = 140 / 1¾ = 80.
b2 = b1 * q = 80 * ¾ = 60.
b3 = b2 * q = 60 * ¾ = 45.
Answer: 80, 60, 45.