jmc

Let $d$ be the common difference, and let $r$ be the common ratio, so $d$ and $r$ are positive integers. Then $a_n=1+(n-1)d$ and $b_n=r^{n-1},$ so $$\begin{array}{rl}1+(k-2)d+r^{k-2}& =100,\\ 1+kd+r^k& =1000.\end{array}$$Then $$\begin{array}{rl}(k-2)d+r^{k-2}& =99,\\ kd+r^k& =999.\end{array}$$From the second equation, $r^k<999.$ If $k\ge 4,$ then $r<999^{1/4},$ so $r\le 5.$

Since the geometric sequence is increasing, $r\ne 1,$ so the possible values of $r$ are 2, 3, 4, and 5. We can write the equations above as $$\begin{array}{rl}(k-2)d& =99-r^{k-2},\\ kd& =999-r^k.\end{array}$$Thus, $99-r^{k-2}$ is divisible by $k-2,$ and $999-r^k$ is divisible by $k.$

If $r=2,$ then the only possible values of $k$ are 4, 5, 6, 7, and 8. We find that none of these values work.

If $r=3,$ then the only possible values of $k$ are 4, 5, and 6. We find that none of these values work.

If $r=4,$ then the only possible values of $k$ is 4. We find that this value does not work.

If $r=4,$ then the only possible values of $k$ is 4. We find that this value does not work.

Therefore, we must have $k=3,$ so $$\begin{array}{rl}d+r& =99,\\ 3d+r^3& =999.\end{array}$$From the first equation, $d=99-r.$ Substituting, we get $$3(99-r)+r^3=999,$$so $r^3-3r-702=0.$ This factors as $(r-9)(r^2+9r+78)=0,$ so $r=9,$ so $d=90.$ Then $a_3=1+2\cdot 90=181$ and $c_3=9^2=81,$ and $c_3=181+81=\boxed{262}.$