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

August 22, 2021 | education

| **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.

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