一个等比数列的第二项为 $16$，无穷和为 $100$。

(a) 求首项的两个可能值。

(b) 证明其中一个可能的等比数列的第 $n$ 项，等于另一个等比数列的第 $n$ 项乘以 $4^{n - 2}$。

用第二项和无穷和列两个方程。消去首项后，先解公比并检查收敛条件。

With first term $a$ and ratio $r$, we have $ar=16$ and $a=100(1-r)$. Thus

Both $r=4/5$ and $r=1/5$ have absolute value below $1$. The corresponding first terms $a=16/r$ are $20$ and $80$.

也可以先消去公比，直接得到关于首项的二次方程。

The condition $ar=16$ gives $r=16/a$. The sum condition becomes $a=100(1-16/a)$, so

Both $a=20$ and $a=80$ give a convergent progression with the required second term.

两列数的首项和公比都已知。将两个通项相除，指数部分的底数正好是 $4$。

The two $n$th terms are $u_n=20(4/5)^{n-1}$ and $v_n=80(1/5)^{n-1}$. Their ratio is

Thus $u_n=4^{n-2}v_n$.