The 42nd question: Given infinite real sequences $a_{1}, a_{2}, \ldots$ and $b_{1}, b_{2}, \ldots$, for $n \geq 1$, we have $a_{n+1}+b_{n+1}=\frac{a_{n}+b_{n}}{2}$, $a_{n+1} b_{n+1}=\sqrt{a_{n} b_{n}}$. Suppose $b_{2016}=1$ and $a_{1}>0$, find all possible values of $a_{1}$.

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The 42nd question: Given infinite real sequences \(a_{1}, a_{2}, \ldots\) and \(b_{1}, b_{2}, \ldots\), for \(n \geq 1\), we have \(a_{n+1} + b_{n+1} = \frac{a_{n} + b_{n}}{2}\) and \(a_{n+1} b_{n+1} = \sqrt{a_{n} b_{n}}\). Suppose \(b_{2016} = 1\) and \(a_{1} > 0\), find all possible values of \(a_{1}\).
Question 42:
Solution: Let $A_{n}=a_{n}+b_{n}, B_{n}=a_{n} b_{n}$, then for any positive integer $n$, we have $\left\{\begin{array}{l}A_{n+1}=\frac{A_{1}}{2} \\ B_{n+1}=B_{n}^{\frac{1}{2}}\end{array}\right.$, and $A_{n}^{2}-$ $4 B_{n}=\left(a_{n}-b_{n}\right)^{2} \geq 0$, which means $A_{n}^{2} \geq 4 B_{n}$.
For any positive integer $n$, according to the recursive relation, we have: $\left\{\begin{array}{c}A_{n}=\frac{A_{1}}{2^{n-1}} \\ B_{n}=B_{1}^{\frac{1}{2^{n-1}}}\end{array}\right.$ Therefore, when $n \rightarrow+\infty$, we have $A_{n} \rightarrow 0$.
If $B_{1} \neq 0$, then when $n \rightarrow+\infty$, we have $B_{n} \rightarrow 1$. However, $B_{n} \leq \frac{A_{n}^{2}}{4} \rightarrow 0$, which leads to a contradiction. Hence, $B_{1}=0$. Consequently, for any positive integer $n$, we have $B_{n}=0$. Thus, $B_{2016}=a_{2016} b_{2016}=0$, which implies $a_{2016}=0$.
Also, $a_{1} b_{1}=B_{1}=0 \Rightarrow b_{1}=0$, hence $A_{1}=a_{1}+b_{1}=a_{1}$. Therefore, $1=a_{2016}+b_{2016}=$ $A_{2016}=\frac{A_{1}}{2^{2015}}=\frac{a_{1}}{2^{2015}}$, so $a_{1}=2^{2015}$.
On the other hand, for any positive integer $n$, let $a_{n}=\left[1+(-1)^{n+1}\right]^{2016-n}, b_{n}=\left[1+(-1)^{n}\right]^{2016-n}$, then the conditions are satisfied.
In summary, the value of $a_{1}=2^{2015}$.