smc

Let our geometric sequence be $a,ar,a{r}^2$ and let our arithmetic sequence be $0,d,2d$. We know that $$\{\begin{array}{l}a+0=1\\ ar+d=1\\ a{r}^2+2d=2\end{array}$$ This implies that $a=1$, hence $r+d=1$ and $r^2+2d=2$. We can rewrite $r+d=1$ as $d=1-r$, plugging this into $r^2+2d=2$, we get $r^2+2-2r=2$, we can simplify this to get $r(r-2)=0$, so $r=0$ or $2$. But since $r\ne 0$, $r=2$ and $d=-1$. So our two sequences are $\{1-n\}$ and $\{2^{n-1}\}$, which means the third sequence will be $$\{2^{n-1}+n+1\}=\{1,1,2,5,12,27,\dots \}$$Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is $\boxed{978}$.