Question

一个等差数列各项之和为 $715$ 。将第一项增加 $1$ ，第二项增加 $3$ ，第三项增加 $5$ ，一般地，将第 $k$ 项增加第 $k$ 个正奇数。新数列各项之和为 $836$ 。求原数列的首项、末项和中间项之和。

The terms of an arithmetic sequence add to $715.$ The first term of the sequence is increased by $1,$ the second term is increased by $3,$ the third term is increased by $5,$ and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836.$ Find the sum of the first, last, and middle terms of the original sequence.

答案

答案：195

知识点：等差数列前n个奇数之和平均数

难度评级：1790

解答：

若数列有 $n$ 项，增加量是前 $n$ 个正奇数之和，即 $n^2$ 。因此 $n^2 = 836 - 715 = 121$ ，所以 $n = 11$ 。

原来 $11$ 项的平均数为 $\frac{715}{11} = 65$ ，这正是等差数列的中间项，也就是第六项。首项和末项的平均数同样为 $65$ ，所以它们的和为 $130$ 。

所求和为 $65 + 130 = 195$ 。

If the sequence has $n$ terms, the amounts added are the first $n$ odd numbers, whose sum is $n^2.$ Thus $n^2 = 836 - 715 = 121,$ so $n = 11.$

The average of the $11$ terms is $\frac{715}{11} = 65,$ which equals the middle (sixth) term of the arithmetic sequence. The first and last terms also average to $65,$ so they add to $130.$

The requested sum is $65 + 130 = 195.$

2012 AIME I 第 2 题

答案：195

难度评级：1790

解答：

其他年份的第 2 题