definition

Generally speaking, if a series starts from the second term, the difference between each term and its previous term is equal to the same constant. This series is called arithmetic progression, and this constant is called arithmetic progression's tolerance. Tolerances are usually represented by the letter d, and the sum of the first n terms is represented by Sn.

abbreviate

A.p. (arithmetic progression can be abbreviated as A.P.).

arithmetic mean

Arithmetic progression, which consists of three numbers A, A and B, can be called the simplest arithmetic progression. At this time, A is called the arithmetic average of A and B, which is very important: A=(a+b)/2.

General term formula

When an = a1+(n-1) da1= s1(n =10), when an = KN+-s (n-1) (n

Sum of the first n terms

The sum formula of the first n items in reverse addition is: Sn = a1+A2+A3+An = A1+(A1+D)+[A++[an-(n-1) D] 22Sn = (.

nature

And the relationship between any two terms am and an is: an=am+(n-m)d, which can be regarded as arithmetic progression's generalized general term formula. From arithmetic progression's definition and general formula, we can also deduce the first n terms and formulas: A1+an = A2+an-1= A3+an-2 = … = AK+an-k+1,k ∈ {1. Then am+an = AP+aqs2n-1= (2n-1) an, s2n+1= (2n+1) (an+1) sk, S2k-Sk, s3k. The sum of the first n terms = (the first term+the last term) × the number of terms ÷2 = (the last term-the first term) ÷ tolerance+1 the first term =2× the number of the first n terms and the last term-the last term =2× the number of the first n terms and the last term-let the first term be A66. Then a2 is the arithmetic average, then 2 times a2 equals a 1+a3, that is, 2a2=a 1+a3.

App application

In daily life, people often use arithmetic progression. For example, when grading the sizes of various products, when the maximum size is not much different from the minimum size, they often grade according to arithmetic progression. If it is arithmetic progression, and an = m and am = n, then a(m+n)=0. Its application in mathematics can be exemplified as follows: there is more than one algorithm to quickly calculate the integer multiple of 6 from 23 to 132. This paper introduces the first term for calculating arithmetic progression a 1=24(24 is 4 times of 6), and the arithmetic difference is d=6. Then let an = 24+(n- 1) * 6.