算术序列公式之和

序列是任何事物或一组数字按一定顺序遵循规则的排列。基本上，它是一组遵循特定模式的数字（或项目）。例如，5、10、15、20……。是一个序列，因为每次值递增 5。如果序列的元素按升序排列，则序列的顺序为升序。如果序列的元素是降序的，则序列的顺序是降序的。算术数列、几何数列、斐波那契数列、谐波数列、三角数列、平方数列和立方数列是特定数列的一些示例。

算术序列

等差数列是一个数列，其中每个后续项是其前一项和一个常数整数的和。这个常数被称为共同差。因此，算术级数中每两个连续项之间的差异是相同的。

If the first term of an arithmetic sequence is a and the common difference is d, then the terms of the arithmetic sequence are of the form:

 a, a+d, a+2d, a+ 3d, a+4d, ….

Suppose n is the total number of terms in the sequence.

For n = 1, the sequence is a.

For n = 2, the sequence is a, a + d.

For n = 3, the sequence is a, a + d, a + 2d.

For n = 4, the sequence is a, a + d, a + 2d, a + 3d.

Hence, the general term of the sequence is a_n = a + (n – 1)d.

等差数列之和

计算等差数列中所有项之和的公式定义为等差数列公式之和。如果一个算术数列写成它的项相加的形式，例如 a + (a+d) + (a+2d) + (a+3d) + .....，那么它被称为算术数列。第 n 项未知的算术级数的前 n 项之和由下式给出：

S_n = n/2 [2a + (n – 1)d]

where,

S_n = sum of the arithmetic sequence,

a  = first term of the sequence,

d = difference between two consecutive terms,

n = number of terms in the sequence.

If we write 2a in the formula as (a + a), the formula becomes, S_n = n/2 [a + a + (n – 1)d]

We know, a + (n – 1)d is denoted by a_n. Hence, the formula becomes, S_n = n/2 [a + a_n]

推导

Suppose the first term of a sequence is a, common difference is d and the number of terms are n.

We know the n^th term of the sequence is given by, 

a_n = a + (n – 1)d         …… (1)

Also the sum of the arithmetic sequence is,

S_n = a + (a + d) + (a + 2d) + (a + 3d) + …… +  a + (n – 1)d     …… (2)

From (1), the equation (2) can also be expressed as,

S_n = a_n + a_n – d + a_n – 2d + a_n – 3d + …… +  a_n – (n – 1)d        …… (3)

Adding (2) and (3) we get,

2 S_n = [a + (a + d) + (a + 2d) + (a + 3d) + …… +  a + (n – 1)d] + [a_n + a_n – d + a_n – 2d + a_n – 3d + …… +  a_n – (n – 1)d]

2 S_n = (a + a + a + ….. n times) + (a_n + a_n + a_n + ….. n times)

2 S_n = n (a + a_n)

S_n = n/2 [a + a_n]

This derives the formula for sum of an arithmetic sequence.

示例问题

问题 1. 求等差数列之和：4, 10, 16, 22, ...... 最多 10 项。

解决方案：

We have, a = 4, d = 10 – 4 = 6 and n = 10.

Use the formula S_n = n/2 [2a + (n – 1)d] to find the required sum.

S₁₀ = 10/2 [2(4) + (10 – 1)6]

= 5 (8 + 54)

= 5 (62)

= 310

问题 2. 求等差数列的总和：7, 9, 11, 13, ...... 最多 15 项。

解决方案：

We have, a = 7, d = 9 – 7 = 2 and n = 15.

Use the formula S_n = n/2 [2a + (n – 1)d] to find the required sum.

S₁₅ = 15/2 [2(7) + (15 – 1)2]

= 15/2 (14 + 28)

= 15/2 (42)

= 315

问题 3. 如果等差数列的和为 240，且 12 项之间的公差为 2，则求该等差数列的第一项。

解决方案：

We have, S = 200, d = 2 and n = 12.

Use the formula S_n = n/2 [2a + (n – 1)d] to find the required value.

=> 200 = 12/2 [2a + (12 – 1)2]

=> 240 = 6 (2a + 22)

=> 40 = 2a + 22

=> 2a = 18

=> a = 9 

问题 4. 求 8 项和为 116 且第一项为 4 的等差数列的公差。

解决方案：

We have, S = 116, a = 4, n = 8.

Use the formula S_n = n/2 [2a + (n – 1)d] to find the required value.

=> 116 = 8/2 [2(4) + (8 – 1)d]

=> 116 = 4 (8 + 7d)

=> 29 = 8 + 7d

=> 7d = 21

=> d = 3

问题 5. 求第一项和最后一项分别为 4 和 10 的 8 项等差数列的和。

解决方案：

We have, a = 4, n = 8 and a_n = 10.

Use the formula S_n = n/2 [a + a_n] to find the required sum.

S₈ = 8/2 [4 + 10]

= 4 (14)

= 56

问题 6. 求第一项、最后一项和总和分别为 16、12 和 140 的等差数列的项数。

解决方案：

We have, S = 140, a = 16 and a_n = 12.

Use the formula S_n = n/2 [a + a_n] to find the required value.

=> 140 = n/2 [16 + 12]

=> 140 = n/2 (28)

=> 14n = 140

=> n = 10

问题 7. 求一个等差数列的第一项、公差和最后一项分别为 8、7 和 50 的和。

解决方案：

We have, a = 8, d = 7 and a_n = 50.

Use the formula a_n = a + (n – 1)d to find n.

=> 50 = 8 + (n – 1)7

=> 42 = 7 (n – 1)

=> n – 1 = 6

=> n = 7

Use the formula S_n = n/2 [a + a_n] to find the sum of sequence.

S₇ = 7/2 (8 + 50)

= 7/2 (58)

= 203