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?

$1$

$4$

$36$

$49$

$81$

Solution:

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d.$ After the additions, the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $(11 + d)^2 = 9(29 + 2d),$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The corresponding third terms $29 + 2d$ are $49$ and $1,$ so the smallest possible value is $1.$

Thus, the correct answer is A.

2004 AMC 12A Problem 14

Solution:

Problem 14 in Other Years