A.P + GP $\neq$ A.G.P

Let the first term of $A.P.$ be ${a}_{0}$ and the first term of $G.P.$ be ${g}_{0}$ and the common difference of the $A.P.$ be $d$ and the common ratio be $r$

Then we have the four equations as :

${a}_{0}+{g}_{0}=0$ (i)

${a}_{0}+d+{g}_{0}{r}=0$ (ii)

${a}_{0}+2d+{g}_{0}{r}^{2}=1$ (iii)

${a}_{0}+3d+{g}_{0}{r}^{3}=0$ (iv)

Using $(i)$ in $(ii),(iv)$ we get :

${a}_{0}+d={a}_{0}{r}_{0}$ (v)

${a}_{0}+3d={a}_{0}{r}^{3}$ (vi)

Multiplying $(v)$ by $3$ and subtracting $(vi)$ from it we get :

$2{a}_{0}={a}_{0}(3r-{r}^{3})$ (vii)

$\Rightarrow {a}_{0}=0$ or ${r}^{3}-3r+2=0$

But ${a}_{0} \neq 0$ since then we won't have all our equations satisfied.

$\Rightarrow {r}^{3}-3r+2=0$

$\Rightarrow r=1 , -2$

But $r \neq 1$ because there also we can't get all the equations satisfied.

Hence $r=-2$

Solving for others we get :

${a}_{0}=\frac{-1}{9} , {g}_{0}=\frac{1}{9} , d=\frac{1}{3} , r=-2$

So our eleven term of our set is :

${a}_{0}+10d+{g}_{0}{r}^{10}=\frac{-1}{9}+\frac{10}{3}+\frac{1024}{9}=117$