This is a beta course, so its structure, chapters, and examples may continue to change.
Properties of Arithmetic Sequences
Besides the basic definition and formulas, arithmetic sequences have many important properties. Mastering these properties allows you to solve problems more flexibly, and even discover some elegant solutions.
From a geometric viewpoint, the terms of an arithmetic sequence are evenly distributed on the number line, and each term lies exactly at the midpoint of its two neighbors.
The Symmetry Property
In the arithmetic sequence {an}, if m+n=p+q (where m,n,p,q are positive integers), then:
Taking the sum of every consecutive k terms of an arithmetic sequence as a term, the new sequence thus formed is still an arithmetic sequence.
Let Sk, S2k−Sk, S3k−S2k, … form the new sequence. This new sequence is an arithmetic sequence with common difference k2d.
Relationship with the Sum of the First n Terms
The Relationship between the General Term and the Sum of the First n Terms
For an arithmetic sequence {an}, the sum Sn of the first n terms and the general term an satisfy:
an={S1,Sn−Sn−1,n=1n≥2
The Quadratic Function Property of the Sum
The sum Sn of the first n terms of an arithmetic sequence is a quadratic function of n (when d=0):
Sn=2dn2+(a1−2d)n
This is a quadratic function with no constant term.
Properties:
When d>0, Sn has a minimum
When d<0, Sn has a maximum
When d=0, Sn=na1 (a linear function)
Worked Examples
Example 1: Using the Symmetry Property
In the arithmetic sequence {an}, given a3+a7=20, find a5.
Solution:
Since 3+7=2×5, by the symmetry property:
a3+a7=2a5
Therefore:
2a5=20⇒a5=10
Example 2: Using the Subsequence Property
The common difference of the arithmetic sequence {an} is d=2. Extract the 1st, 3rd, 5th, … terms to form a new sequence {bn}. Find the common difference of {bn}.
Solution:
b1=a1, b2=a3, b3=a5, …
b2−b1=a3−a1=2d=2×2=4
So the common difference of the new sequence is 4.
Example 3: Using the Sum Property
The sum of the first n terms of the arithmetic sequence {an} is Sn=2n2+3n. Find a1 and d.
Solution:
When n=1:
a1=S1=2×12+3×1=5
When n≥2:
an=Sn−Sn−1=(2n2+3n)−[2(n−1)2+3(n−1)]=2n2+3n−2n2+4n−2−3n+3=4n+1
Verify n=1: a1=4×1+1=5 ✓
So an=4n+1, i.e., a1=5 and d=4.
Practice Problems
Exercise 1
In the arithmetic sequence {an}, a4+a6=18. Find a5.
Reference Answer(2 个标签)
arithmetic sequencesymmetry property
Idea: Use the symmetry property.
Detailed steps:
Since 4+6=2×5, we have:
a4+a6=2a5
18=2a5
a5=9
Answer: a5=9
Exercise 2
In the arithmetic sequence {an}, a2+a4+a6+a8+a10=80. Find a6.
Reference Answer(2 个标签)
arithmetic sequencesymmetry property
Idea: Use the symmetry property and the middle-term property.