jmc

Let the arithmetic sequence be $a_n=a+(n-1)d,$ and let the geometric sequence be $b_n=b{r}^{n-1}.$ Then $$\begin{array}{rl}a+b& =1,\\ a+d+br& =4,\\ a+2d+b{r}^2& =15,\\ a+3d+b{r}^3& =2.\end{array}$$Subtracting pairs of equations, we get $$\begin{array}{rl}d+br-b& =3,\\ d+b{r}^2-br& =11,\\ d+b{r}^3-b{r}^2& =-13.\end{array}$$Again subtracting pairs of equations, we get $$\begin{array}{rl}b{r}^2-2br+b& =8,\\ b{r}^3-2b{r}^2+br& =-24.\end{array}$$We can write these as $$\begin{array}{rl}b(r-1{)}^2& =8,\\ br(r-1{)}^2& =-24.\end{array}$$Dividing these equations, we get $r=-3.$ Then $16b=8,$ so $b=\frac{1}{2}.$ Then $$\begin{array}{rl}a+\frac{1}{2}& =1,\\ a+d-\frac{3}{2}& =4.\end{array}$$Solving for $a$ and $d,$ we find $a=\frac{1}{2}$ and $d=5.$

Hence, $$\begin{array}{rl}{c}_5& =a_5+b_5\\ & =a+4d+b{r}^4\\ & =\frac{1}{2}+4\cdot 5+\frac{1}{2}\cdot (-3{)}^4\\ & =\boxed{61}.\end{array}$$