一个等比数列的前三项分别为 $a$、$b$ 和 $c$，其中 $a$、$b$ 和 $c$ 是正常数。一个等差数列的前三项分别为 $a$、$b$ 和 $-3c$。

(a) 证明 $a^2-10ac+9c^2=0$。

现在给定 $a=9$，并且 $c$ 取两个可能值中较小的一个。

(b) (i) 求该等比数列的无穷和。

(ii) 求该等差数列前 20 项的和。

等差条件给出中项与两个端点的和，等比条件给出中项的平方。把两个条件合起来消去 $b$。

The arithmetic progression gives $b-a=-3c-b$, or $2b=a-3c$. The geometric progression gives $b/a=c/b$, so $b^2=ac$. Squaring the first relation and substituting into it gives

Rearranging gives $a^2-10ac+9c^2=0$.

代入 $a=9$，再把二次方程的根代回 $2b=a-3c$，检查 $b>0$。由首项和第三项求公比，并检查无穷和的收敛条件。

With $a=9$, the equation in (a) becomes

When $c=9$, the original arithmetic progression condition $2b=9-3c$ gives $b=-9$, outside the given domain. When $c=1$, it gives $b=3$, which also satisfies $b^2=9c$. Thus $c=1$. For the common ratio $r$, $9r^2=1$, and $r>0$ because the terms are positive. Hence $r=\frac13$. Since $|r|<1$,

直接用等比数列的公比表示 $b$ 和 $c$，再用等差数列的条件解公比。

Let $r>0$ be the common ratio. Then $b=9r$ and $c=9r^2$. The arithmetic progression condition $2b=9-3c$ gives

Only $r=\frac13$ is positive. It gives $b=3$ and $c=1$, and the arithmetic terms $9,3,-3$ do have a common difference. Since $r<1$, the geometric sum to infinity is $9/(1-\frac13)=\frac{27}2$.

两列数的首两项相同。用 (i) 的公比找出第二项，就可以计算等差数列的公差。

From (i), $b=9\cdot\frac13=3$. Hence the arithmetic progression has first term $9$ and common difference $3-9=-6$. Its first $20$ terms sum to