一个收敛等比数列的首项为 10。该数列前 4 项的和为 $p$，前 8 项的和为 $q$。已知 $\frac{q}{p} = \frac{17}{16}$。

求无穷和的两个可能值。

用有限等比数列的求和公式表示 $p$ 和 $q$。两式相除后，可以用平方差分解约掉 $1-r^4$。

Let the common ratio be $r$. Convergence gives $|r|<1$, so neither $1-r$ nor $1-r^4$ is zero. The finite sums give

Thus $1+r^4=\frac{17}{16}$ and $r^4=\frac1{16}$. Both real ratios $r=\pm\frac12$ give convergent progressions. The sum to infinity is $10/(1-r)$, giving $20$ for $r=\frac12$ and $\frac{20}3$ for $r=-\frac12$.

前八项可分成前四项和接下来的四项。后一组是前一组逐项乘 $r^4$ 得到的，这样可以直接用比值求公比。

Let the common ratio be $r$, with $|r|<1$ by convergence. Terms 5 to 8 have sum $r^4p$, so

It follows that $r^4=\frac1{16}$ and $r=\frac12$ or $r=-\frac12$. Both ratios satisfy $|r|<1$, and the corresponding sums to infinity are