如何找到给定总和的算术级数的公差？
序列被定义为按顺序排列的项目或对象的列表。它也被定义为遵循某些规则的已定义顺序的一组数字。假设如果a 1 , a 2 , a 3 , a 4 ... 等表示给定序列的项，则下标 1, 2, 3, 4... 表示序列项的位置。
有几种类型的序列，其中一些是：
- 算术序列
- 几何序列
- 谐波序列
算术数列是一个数列，其中每一项都是通过对前面的数加上或减去一个常数来创建的。例如，整数系列是 0, 1, 2, 3, 4,… 是一个公差为 1 的算术级数。
在算术级数( AP ) 中，有三个主要项。这些被表示为：
- 共同点差（ d ）
- 第 n项 ( a n )
- AP 总和 ( S n )
两个连续数字之间的差异称为共同差异。考虑AP中的数字a 1 ， a 2 ， a 3 ， a 4 ... a n ，共同差由以下表达式计算：
d = a2 – a1, a3 – a2, a4 – a3…, an – an-1
The AP can also be written in the following form:
a, a+d, a+2d, a+3d,… … …, a+(n-1)d
where a is the first term of the progression, d is the common difference of the progression and a+(n-1)d is the nth term of AP.
The sum of the AP is given by
S = n/2[2a + (n – 1)d]
给定总和时计算公差的步骤：
计算公差的步骤如下：
- 代入 sum 的值、项数和公式中的第一项。
- 简化右手边。
- 求解d的值。
- 确保计算正确。
考虑AP的第一项是5，AP的项数是9，AP的和是189。计算上述算术级数的公差。
解决方案：
Given that,
The first term of the AP is 5,
The number of terms in the AP is 9 and
The sum of the AP is 189.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 9 for n, 5 for a and 189 for Sn into the formula.
189=9/2[2×5+(9-1)d]
42=10+8×d
8d=32
d=4
Hence, the common difference of the given series is 4.
类似问题
问题1：假设一个AP的第一项是120，AP的项数是5，AP的总和是650。计算上述算术级数的公差。
解决方案：
Given that,
The first term of the AP is 120,
The number of terms in the AP is 5 and
The sum of the AP is 650.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 5 for n, 120 for a and 650 for Sn into the formula.
650 = 5/2[2 × 120 + (5-1)d]
260 = 240 + 4 × d
4d = 20
d = 5
Hence, the common difference of the given series is 5.
问题2：假设一个AP的第一项是5.5，AP的项数是10，AP的和是100。计算上述算术级数的公差。
解决方案：
Given that,
The first term of the AP is 5.5,
The number of terms in the AP is 10 and
The sum of the AP is 100.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 10 for n, 5.5 for a and 100 for Sn into the formula.
100 = 10/2[2 × 5.5 + (10-1)d]
20 = 11 + 9 × d
9d = 9
d = 1
Hence, the common difference of the given series is 1.