= Solution
Apply the method to the <Dahlquist test equation>, put $z=h\lambda$, and retain every root of its <amplification polynomial of a multistep method>:
$$
P_z(w)=\left[1-\frac{1+a}{2}z\right]w^2
-\left[1+a+\frac{1-3a}{2}z\right]w+a.
$$
An <A-stable> convergent method must satisfy the <root condition for a multistep method> for every $\operatorname{Re}z\leq0$.
First take $-1<a<0$. As negative real $z$ tends to minus infinity, one root tends to the nonzero root of $\sigma$,
$$
w_\infty=\frac{3a-1}{1+a},\qquad |w_\infty|>1.
$$
It follows by root continuity that some sufficiently negative finite $z$ is unstable. At $a=-1$, the recurrence has <polynomial> $w^2-2zw-1$; for any negative real $z$, its root $z-\sqrt{z^2+1}$ is below $-1$. Thus all negative convergent parameters are excluded.
For $0\leq a<1$, use the <boundary-locus test for multistep A-stability>. A unit root $w=e^{i\theta}$ with $\sigma(w)\ne0$ would require $z=\rho(w)/\sigma(w)$. Direct multiplication by the conjugate denominator gives
$$
\operatorname{Re}\frac{\rho(e^{i\theta})}{\sigma(e^{i\theta})}
=\frac{a(1+a)(1-\cos\theta)^2}{|\sigma(e^{i\theta})|^2}\geq0.
$$
There is therefore no unit-root crossing in the open left half-plane. The leading coefficient of $P_z$ never vanishes there, since its only zero is the positive real value $2/(1+a)$. For a small negative real $z$, the simple root initially at one moves to $1+z+O(z^2)$, inside the disk, and the other root is close to $a$, also inside. Continuity of <polynomial> roots and connectedness of the open left half-plane then keep both roots strictly inside throughout that half-plane. Possible zeros of $\sigma$ cause no omitted crossing: for $a>0$ its other root lies strictly inside the <unit disk>, while at $a=0$ its unit-circle zero $w=-1$ has $\rho(-1)\ne0$.
On the imaginary axis, if $a>0$ the displayed real part vanishes only at $w=1$, which corresponds to $z=0$ and is simple. Other imaginary-axis parameters retain strictly interior roots. If $a=0$, the two roots are zero and the <stability function> $(1+z/2)/(1-z/2)$, whose modulus is one on that axis; the unit root is simple. Thus the required boundary <root condition for a multistep method> holds as well.
Combining necessity and sufficiency gives
$$
\boxed{\text{The convergent A-stable members are exactly }0\leq a<1.}
$$
In particular, the third-order member $a=-1/5$ is convergent but not <A-stable>, in agreement with the <Second Dahlquist barrier>.
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