(i) (a) 一个等差数列的首项为 $a$，公差为 $d$。证明：(3)

(b) 由此求以下和。(3)

(ii) 已知一个等比数列的前三项为

其中 $k$ 为常数。

(a) 证明：(3)

又已知该级数收敛。

(b) 求 $S_\infty$ 的值。(4)

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:

Each paired column sums to $2a+(n-1)d$, and there are $n$ such pairs. 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),

等比数列满足 $u_2^2=u_1u_3$，代入前三项后展开整理。

For a geometric series, $u_2^2=u_1u_3$, so

Expanding both sides,

解二次方程求 $k$，再根据 $|r|<1$ 选择收敛的公比，最后用 $S_\infty=\dfrac{a}{1-r}$。

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