Fixed point iteration is sometimes used for solving the implicit Euler method
$$y_{n+1}^\mathrm{(I)}= y_n + hf(y_{n+1}^\mathrm{(I)})$$
with the iteration procedure starting with explicit Euler
$$y_{n+1}^\mathrm{(E)}= y_n + hf(y_n)$$
My question is, once we achieve convergence on the iteration and we have both the value computed implicitly and the explicit value that we used to start the iteration, cannot we combine both values to get a more accurate result?
$$y_{n+1}^\mathrm{(A)} := \frac{y_{n+1}^\mathrm{(I)} + y_{n+1}^\mathrm{(E)}}{2} = y_n + h \frac{f(y_{n+1}^\mathrm{(I)}) + f(y_n)}{2}$$
because the above average is similar to the trapezoidal method
$$y_{n+1}^\mathrm{(T)}= y_n + h \frac{f(y_{n+1}^\mathrm{(T)}) + f(y_n)}{2}$$