= Solution
For this <two-step family with a third-order member>, use the <exponential-symbol order criterion for a multistep method>. Expanding the exact-solution residual symbol gives
$$
\rho(e^z)-z\sigma(e^z)
=-\frac{1+5a}{12}z^3-\frac{3+11a}{24}z^4+O(z^5).
$$
The constant, linear and quadratic coefficients vanish for every parameter. Thus
$$
\boxed{p=2\text{ for }a\ne-1/5,\qquad p=3\text{ for }a=-1/5.}
$$
At the exceptional value the fourth-degree coefficient is $-1/30$, so there is no further increase. This means an unscaled exact-solution residual of order $h^{p+1}$, or a residual divided by the step of order $h^p$.
The first characteristic <polynomial> factors as $\rho(w)=(w-1)(w-a)$. The <root condition for a multistep method> requires every root to have modulus at most one and each unit-modulus root to be simple. The root $a=-1$ is allowed because it is distinct from the simple root one; $a=1$ is not, since it makes one a double root. Hence <zero-stability> holds exactly for
$$
\boxed{-1\leq a<1.}
$$
On this interval, $\rho(1)=0$ and $\rho'(1)=\sigma(1)=1-a\ne0$ establish <consistency of a numerical method>. The <Dahlquist equivalence theorem>, for a fixed-step consistent <linear multistep method> applied to an <ordinary differential equation> with a <Lipschitz continuous> right-hand side with convergent starting values and well-defined small-step updates, says that <convergence of a numerical method> is equivalent to <zero-stability>. Therefore the same interval is exactly the <convergence of a numerical method> range. With sufficiently accurate starting values the convergent members have the orders just computed.
At $a=1$ the <Taylor expansion> still gives a formal order-two residual, but $\rho'(1)=\sigma(1)=0$ and the double root prevents a convergent method. Canceling the common factor $w-1$ produces the <common-factor cancellation defect in a multistep recurrence> and changes the starting relations and cannot repair <convergence of a numerical method> of the original recurrence. At $a=0$, cancellation of a zero root instead reveals the ordinary <trapezoidal rule>; it is consistent with the order-two result.
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