一个等差数列的首项为 $1.5$，前十项和为 $127.5$。

(a) 求公差。

(b) 求该等差数列中所有数值在 $25$ 和 $100$ 之间的项的和。

把首项、前十项和及项数代入等差求和公式。

The equation $\frac{10}2(2(1.5)+9d)=127.5$ gives $3+9d=25.5$. Thus the common difference is $d=2.5$.

用通项不等式找出符合范围的整数项号。把这段连续项看成一个新的等差数列。

The $n$th term is $1.5+2.5(n-1)=2.5n-1$. The condition $25<2.5n-1<100$ gives $10.4<n<40.4$, so the indices are $11$ through $40$. There are $30$ terms, starting at $26.5$ and ending at $99$. Their sum is

也可以求出截至第 $40$ 项的总和，再减去前 $10$ 项。

The first $40$ terms sum to $\frac{40}2(1.5+99)=2010$. The first $10$ terms sum to $127.5$. Their difference is $2010-127.5=1882.5$.