jmc

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 of the geometric progression?

Accepted Answer

The terms of the arithmetic progression are 9, 9+d, and 9+2d for some real number d. The terms of the geometric progression are 9, 11+d, and 29+2d. Therefore (11+d)^2 = 9(29+2d) so d^2+4d-140 = 0. Thus d=10 or d=-14. The corresponding geometric progressions are 9, 21, 49 and 9, -3, 1, so the smallest possible value for the third term of the geometric progression is 1.