jmc

Since $\{a_n\}$ is an arithmetic sequence, we may let $a_n=a+(n-1)d$ for some $a$ and $d.$ Since $\{g_n\}$ is a geometric sequence, we may let $g_n=c{r}^{n-1}$ for some $c$ and $r.$ Then we have $$\begin{array}{rl}a+c& =0\\ a+d+cr& =0\\ a+2d+c{r}^2& =1\\ a+3d+c{r}^3& =0.\end{array}$$The first equation gives $c=-a,$ so the remaining equations become $$\begin{array}{rl}a+d-ar& =0\\ a+2d-a{r}^2& =1\\ a+3d-a{r}^3& =0.\end{array}$$From the equation $a+d-ar=0,$ we get $d=ar-a,$ and substituting in the remaining two equations gives $$\begin{array}{rl}-a+2ar-a{r}^2& =1\\ -2a+3ar-a{r}^3& =0.\end{array}$$The equation $-2a+3ar-a{r}^3=0$ factors as $$a(r-1{)}^2(r+2)=0.$$Having $a=0$ would contradict the equation $-a+2ar-a{r}^2=1,$ so either $r=1$ or $r=-2.$ But if $r=1,$ then $\{g_n\}$ is a constant sequence, which means that $\{a_n+g_n\}$ is itself an arithmetic sequence; this is clearly impossible, because its first four terms are $0,0,1,0.$ Thus, $r=-2.$ Then we have $$-a+2a(-2)-a(-2{)}^2=1,$$or $-9a=1,$ so $a=-\frac{1}{9}.$ Then $c=-a=\frac{1}{9}$ and $d=ar-a=-3a=\frac{1}{3}.$ We conclude that $$\begin{array}{rl}{a}_n& =-\frac{1}{9}+(n-1)\frac{1}{3},\\ g_n& =\frac{1}{9}(-2{)}^n\end{array}$$for all $n.$ Then $$a_5+g_5=-\frac{1}{9}+4\cdot \frac{1}{3}+\frac{1}{9}(-2{)}^4=\boxed{3}.$$