标准推导：正序写一遍 $S_n$，倒序写一遍，两式相加。

Writing the sum forwards and backwards:

Adding these two expressions,

识别首项 $a=900$，公差 $d=-8$，末项 $500$。先用通项公式求项数 $n$，再用 (a) 的公式求和。

Here $a=900$ and $d=-8$. The $n$th term is

Setting $u_n=500$:

Using the formula from (a),

等比数列相邻项的比值相等：$\frac{u_2}{u_1}=\frac{u_3}{u_2}$，交叉相乘后展开整理。

For a geometric series, $\dfrac{u_2}{u_1}=\dfrac{u_3}{u_2}$, so

Expanding both sides,

解二次方程求 $k$，再判断哪个 $k$ 使公比 $|r|<1$（收敛条件），最后用 $S_\infty=\frac{a}{1-r}$。

Solving $2k^2-11k-40=0$:

For $k=8$: the first three terms are $12,6,3$, giving $r=\dfrac{6}{12}=\dfrac{1}{2}$. Since $|r|<1$, the series converges.

For $k=-\dfrac{5}{2}$: the first three terms are $\dfrac{3}{2},-\dfrac{9}{2},\dfrac{27}{2}$, giving $r=-3$. Since $|r|>1$, the series diverges.

So $k=8$, $a=12$, $r=\dfrac{1}{2}$. Therefore