本题必须写出所有计算步骤。完全依赖计算器技术的解答不予接受。

一个等差级数的第六项为 $2$，前十项之和为 $-80$。

(a) 求该级数的首项和公差。（4分）

(b) 因此，求使 $S_n>8000$ 成立的最小 $n$ 值。（3分）

设首项为 $a$，公差为 $d$。第六项给一个方程，前十项和再给一个方程，联立即可。

The sixth term is $2$, so

Also,

Thus

Solve

Doubling the first equation gives

Subtract:

Therefore,

以第六项 $2$ 为基准，首项比它少五个公差，第十项比它多四个公差。先用首末项求前十项之和，再求首项。

Since $u_6=2$,

Using $S_{10}=\frac{10}{2}(u_1+u_{10})$,

Therefore,

The first term is $-98$ and the common difference is $20$.

把 $a=-98$，$d=20$ 代入 $S_n$ 公式，解不等式。最后要取最小整数 $n$。

Using

we need

Solve

By the quadratic formula,

The roots are approximately $-23.395$ and $34.195$. Since the quadratic has positive leading coefficient, it is positive outside these roots. As $n$ is a positive integer, we need $n>34.195\ldots$.

Therefore, the smallest integer value of $n$ is $35$.

试错法（Trial and improvement）。由于我们只需要求出满足 $S_n > 8000$ 的最小整数 $n$，代入 $a=-98$ 和 $d=20$ 得到 $S_n = 10n^2 - 108n$。利用计算器直接测试邻近的整数值，找到分界点即可。

From part (a), we have $a = -98$ and $d = 20$.

The sum of the first $n$ terms is given by:

We want to find the smallest integer $n$ such that $S_n > 8000$.

Let’s test integer values for $n$ using a calculator:

• For $n = 34$:

• For $n = 35$:

For $1\leq n\leq5$, $S_n=n(10n-108)<0$. For $n\geq6$, the newly added term $u_n=20n-118$ is positive, so the sums increase.

Since $S_{34} < 8000$ and $S_{35} > 8000$, the smallest integer value of $n$ for which the sum exceeds $8000$ is $35$.