= Solution
For this <Adams–Moulton method>, the <characteristic polynomials of a linear multistep method> are
$$
\rho(w)=w^2-w,
\qquad
\sigma(w)=\frac{5w^2+8w-1}{12}.
$$
Expansion at zero gives
$$
\rho(e^z)-z\sigma(e^z)=-\frac1{24}z^4+O(z^5).
$$
Part (a) therefore shows that the method has order exactly three.
It is consistent because its order is at least one. The roots of $\rho(w)=w(w-1)$ are $0$ and $1$; both lie in the closed unit disk, and the only unit-modulus root is simple. Thus the method is <zero-stability>[zero-stable]. By the <Dahlquist equivalence theorem>, consistency and zero stability imply convergence. Hence the method is
$$
\boxed{\text{third-order and convergent}}.
$$
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