A [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) of three real numbers forms an [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Geometric_progression). What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$
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](https://artofproblemsolving.com/wiki/index.php/Geometric_mean) of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$. The smallest possible value occurs when $d = -14$, and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$.