A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$, $60$, and $91$. What is the fourth term of this sequence?
$\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206$
Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$
We are given that
\begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*}
and we wish to find $a+3d+br^3.$
Subtracting the first equation from the second and the second equation from the third, we get
\begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*}
Subtract these results, we get \[b(r-1)^2=28.\]
Note that either $b=28$ or $b=7.$ We proceed with casework:

If $b=28,$ then $r=2,a=29,$ and $d=-25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction.
If $b=7,$ then $r=3,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ This case is valid.
Therefore, The answer is $a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.$
~mathboy282 ~MRENTHUSIASM