We have
$$ \underbrace{\text{QM}(a_1,\ldots,a_n)}_{Q(n)}\geq \underbrace{\text{AM}(a_1,\ldots,a_n)}_{A(n)} \geq \underbrace{\text{QM}(a_1,\ldots,a_{n+1})}_{Q(n+1)}\geq \underbrace{\text{AM}(a_1,\ldots,a_{n+1})}_{A(n+1)} \tag{0}$$
so both $A(n)$ and $Q(n)$ are non-increasing and $a_{n+1}\leq A(n)$. The central inequality can be written as
$$ a_{n+1}^2 \leq (n+1)A(n)^2 - nQ(n)^2 \tag{1} $$
so we must have
$$ A(n)^2 \geq \frac{n}{n+1} Q(n)^2,\qquad A(n)\geq Q(n)\sqrt{1-\tfrac{1}{n+1}}.$$
We may consider that the average value of $a_1,\ldots,a_n$ is $A(n)$ and
$$ V(n)=\frac{1}{n}\sum_{k=1}^{n}(a_k-A(n))^2 = Q(n)^2-A(n)^2\leq \frac{Q(n)^2}{n+1}.$$
$Q(n)$ is non-increasing and $\frac{1}{n+1}$ is decreasing to zero, so the variance goes to zero as $n\to +\infty$.
We may write $(1)$ as
$$ a_{n+1}^2 \leq A(n)^2 - nV(n) \tag{2}$$
and define a sequence in the following way:
$$ a_1=2,\quad a_2=1,\quad a_{n+1}=\sqrt{A(n)^2-nV(n)} $$
leading to
$$ \{a_n\}_{n\geq 1}=\left\{2,1,\frac{\sqrt{7}}{2},\frac{1}{6} \sqrt{48 \sqrt{7}-71},\frac{1}{12} \sqrt{\frac{15}{2} \sqrt{979+1212 \sqrt{7}}+3 \sqrt{7}-293},\ldots\right\} $$
This seems to work for a few terms, but at some point $n V(n)=\sum_{k=1}^{n}(a_k-A(n))^2$ becomes larger than $A(n)^2$. Now we have to prove that unless $\{a_n\}_{n\geq 1}$ is constant we cannot avoid this phenomenon.
$$\begin{eqnarray*} (n+1)V(n+1)-n V(n) &=& (n+1)Q(n+1)^2-(n+1)A(n+1)^2-n Q(n)^2+n A(n)^2\\&=&(a_{n+1}-A(n+1))^2+n(A(n)-A(n+1))^2\end{eqnarray*} $$
shows that $n V(n)$ is weakly increasing.
$$ (n+1)V(n+1)=\sum_{k=1}^{n}((k+1)V(k+1)-k V(k))\geq \sum_{k=1}^{n}k(A(k)-A(k+1))^2 $$
and
$$n\sum_{k=1}^{n}k(A(k)-A(k+1))^2\stackrel{\text{CS}}{\geq}\left(\sum_{k=1}^{n}\sqrt{k}(A(k)-A(k+1))\right)^2 $$
can be lower-bounded by using summation by parts:
$$ \sum_{k=1}^{n}\sqrt{k}(A(k)-A(k+1)) \geq (A(1)-A(n+1))\sqrt{n}. $$
