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Question

A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

Accepted Answer

Let d be the common difference. Then 9, 9+d+2=11+d, 9+2d+20=29+2d are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, (11+d)^2 = 9(2d+29) d^2 + 4d - 140 = (d+14)(d-10) = 0. The smallest possible value occurs when d = -14, and the third term is 2(-14) + 29 = 11.