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.

The Middle-Term Property

The Arithmetic Mean

Arithmetic mean

If three numbers aa, bb, cc form an arithmetic sequence, then bb is called the arithmetic mean of aa and cc.

Arithmetic mean formula
b=a+c2b = \frac{a + c}{2}

In other words, the arithmetic mean is the arithmetic average of two numbers.

Example: 22, 55, 88 form an arithmetic sequence, and 55 is the arithmetic mean of 22 and 88, because 5=2+825 = \frac{2 + 8}{2}.

The General Middle-Term Property

In the arithmetic sequence {an}\{a_n\}, for any positive integer nn (n2n \geq 2):

Middle-term property
an=an1+an+12a_n = \frac{a_{n-1} + a_{n+1}}{2}

This shows that in an arithmetic sequence, any term is the arithmetic mean of its two neighbors.

Why is every term of an arithmetic sequence the average of its two neighbors?

This follows from the definition of an arithmetic sequence. Let the common difference be dd. Then:

  • an1=anda_{n-1} = a_n - d
  • an+1=an+da_{n+1} = a_n + d

Therefore: an1+an+12=(and)+(an+d)2=2an2=an\frac{a_{n-1} + a_{n+1}}{2} = \frac{(a_n - d) + (a_n + d)}{2} = \frac{2a_n}{2} = a_n

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}\{a_n\}, if m+n=p+qm + n = p + q (where m,n,p,qm, n, p, q are positive integers), then:

Symmetry property
am+an=ap+aqa_m + a_n = a_p + a_q

In particular, when m+n=2pm + n = 2p, we have am+an=2apa_m + a_n = 2a_p.

Example: In an arithmetic sequence, a2+a8=a3+a7=a4+a6=2a5a_2 + a_8 = a_3 + a_7 = a_4 + a_6 = 2a_5 (because 2+8=3+7=4+6=2×5=102+8=3+7=4+6=2 \times 5=10)

Proof:

am+an=[a1+(m1)d]+[a1+(n1)d]=2a1+(m+n2)dap+aq=[a1+(p1)d]+[a1+(q1)d]=2a1+(p+q2)d\begin{aligned} a_m + a_n &= [a_1 + (m-1)d] + [a_1 + (n-1)d] \\ &= 2a_1 + (m+n-2)d \\ \\ a_p + a_q &= [a_1 + (p-1)d] + [a_1 + (q-1)d] \\ &= 2a_1 + (p+q-2)d \end{aligned}

Since m+n=p+qm + n = p + q, we have am+an=ap+aqa_m + a_n = a_p + a_q.

The Subsequence Property

Extracting at Equal Intervals

Extracting several terms from an arithmetic sequence at equal intervals produces a new sequence that is still an arithmetic sequence.

Example: The sequence {an}\{a_n\}: 1,3,5,7,9,11,13,1, 3, 5, 7, 9, 11, 13, \ldots (common difference 2)

Extracting odd-indexed terms: a1,a3,a5,a_1, a_3, a_5, \ldots i.e., 1,5,9,1, 5, 9, \ldots (common difference 4)

Extracting even-indexed terms: a2,a4,a6,a_2, a_4, a_6, \ldots i.e., 3,7,11,3, 7, 11, \ldots (common difference 4)

Sums of Consecutive k Terms

Taking the sum of every consecutive kk terms of an arithmetic sequence as a term, the new sequence thus formed is still an arithmetic sequence.

Let SkS_k, S2kSkS_{2k} - S_k, S3kS2kS_{3k} - S_{2k}, \ldots form the new sequence. This new sequence is an arithmetic sequence with common difference k2dk^2d.

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}\{a_n\}, the sum SnS_n of the first nn terms and the general term ana_n satisfy:

an={S1,n=1SnSn1,n2a_n = \begin{cases} S_1, & n = 1 \\ S_n - S_{n-1}, & n \geq 2 \end{cases}

The Quadratic Function Property of the Sum

The sum SnS_n of the first nn terms of an arithmetic sequence is a quadratic function of nn (when d0d \neq 0):

Sn=d2n2+(a1d2)nS_n = \frac{d}{2}n^2 + \left(a_1 - \frac{d}{2}\right)n

This is a quadratic function with no constant term.

Properties:

  • When d>0d > 0, SnS_n has a minimum
  • When d<0d < 0, SnS_n has a maximum
  • When d=0d = 0, Sn=na1S_n = na_1 (a linear function)

Worked Examples

Example 1: Using the Symmetry Property

In the arithmetic sequence {an}\{a_n\}, given a3+a7=20a_3 + a_7 = 20, find a5a_5.

Solution:

Since 3+7=2×53 + 7 = 2 \times 5, by the symmetry property:

a3+a7=2a5a_3 + a_7 = 2a_5

Therefore: 2a5=20a5=102a_5 = 20 \quad \Rightarrow \quad a_5 = 10

Example 2: Using the Subsequence Property

The common difference of the arithmetic sequence {an}\{a_n\} is d=2d = 2. Extract the 1st, 3rd, 5th, … terms to form a new sequence {bn}\{b_n\}. Find the common difference of {bn}\{b_n\}.

Solution:

b1=a1b_1 = a_1, b2=a3b_2 = a_3, b3=a5b_3 = a_5, \ldots

b2b1=a3a1=2d=2×2=4b_2 - b_1 = a_3 - a_1 = 2d = 2 \times 2 = 4

So the common difference of the new sequence is 44.

Example 3: Using the Sum Property

The sum of the first nn terms of the arithmetic sequence {an}\{a_n\} is Sn=2n2+3nS_n = 2n^2 + 3n. Find a1a_1 and dd.

Solution:

When n=1n = 1: a1=S1=2×12+3×1=5a_1 = S_1 = 2 \times 1^2 + 3 \times 1 = 5

When n2n \geq 2: an=SnSn1=(2n2+3n)[2(n1)2+3(n1)]a_n = S_n - S_{n-1} = (2n^2 + 3n) - [2(n-1)^2 + 3(n-1)] =2n2+3n2n2+4n23n+3=4n+1= 2n^2 + 3n - 2n^2 + 4n - 2 - 3n + 3 = 4n + 1

Verify n=1n = 1: a1=4×1+1=5a_1 = 4 \times 1 + 1 = 5

So an=4n+1a_n = 4n + 1, i.e., a1=5a_1 = 5 and d=4d = 4.

Practice Problems

Exercise 1

In the arithmetic sequence {an}\{a_n\}, a4+a6=18a_4 + a_6 = 18. Find a5a_5.

Reference Answer(2 个标签)
arithmetic sequencesymmetry property

Idea: Use the symmetry property.

Detailed steps:

Since 4+6=2×54 + 6 = 2 \times 5, we have:

a4+a6=2a5a_4 + a_6 = 2a_5

18=2a518 = 2a_5

a5=9a_5 = 9

Answer: a5=9a_5 = 9

Exercise 2

In the arithmetic sequence {an}\{a_n\}, a2+a4+a6+a8+a10=80a_2 + a_4 + a_6 + a_8 + a_{10} = 80. Find a6a_6.

Reference Answer(2 个标签)
arithmetic sequencesymmetry property

Idea: Use the symmetry property and the middle-term property.

Detailed steps:

Since 2+10=4+8=2×62+10=4+8=2 \times 6, we have:

a2+a10=2a6a_2 + a_{10} = 2a_6 a4+a8=2a6a_4 + a_8 = 2a_6

Therefore: a2+a4+a6+a8+a10=2a6+2a6+a6=5a6=80a_2 + a_4 + a_6 + a_8 + a_{10} = 2a_6 + 2a_6 + a_6 = 5a_6 = 80

a6=16a_6 = 16

Answer: a6=16a_6 = 16

Exercise 3

The sum of the first nn terms of the arithmetic sequence {an}\{a_n\} is Sn=n2nS_n = n^2 - n. Find ana_n.

Reference Answer(2 个标签)
arithmetic sequencesum of first n terms

Idea: Use an=SnSn1a_n = S_n - S_{n-1} (n2n \geq 2).

Detailed steps:

When n=1n = 1: a1=S1=121=0a_1 = S_1 = 1^2 - 1 = 0

When n2n \geq 2:

an=SnSn1=(n2n)[(n1)2(n1)]=n2n(n22n+1n+1)=n2nn2+3n2=2n2\begin{aligned} a_n &= S_n - S_{n-1} \\ &= (n^2 - n) - [(n-1)^2 - (n-1)] \\ &= n^2 - n - (n^2 - 2n + 1 - n + 1) \\ &= n^2 - n - n^2 + 3n - 2 \\ &= 2n - 2 \end{aligned}

Verify n=1n = 1: a1=2×12=0a_1 = 2 \times 1 - 2 = 0

Answer: an=2n2a_n = 2n - 2

Exercise 4

Adapted from a postgraduate entrance examination problem

In the arithmetic sequence {an}\{a_n\}, a1>0a_1 > 0 and S9=S12S_9 = S_{12}. After how many terms does the sum reach its maximum?

Reference Answer(2 个标签)
arithmetic sequencesum of first n terms

Idea: S9=S12S_9 = S_{12} implies that the sum of the 10th through 12th terms is 0.

Detailed steps:

S12S9=a10+a11+a12=0S_{12} - S_9 = a_{10} + a_{11} + a_{12} = 0

Since it is an arithmetic sequence: 3a11=0a11=03a_{11} = 0 \quad \Rightarrow \quad a_{11} = 0

Also, since a1>0a_1 > 0, the sequence is decreasing (d<0d < 0).

Therefore:

  • a10>0a_{10} > 0
  • a11=0a_{11} = 0
  • a12<0a_{12} < 0

The sum of the first 10 terms is maximal.

Answer: The sum of the first 10 terms is maximal


Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
ana_n元素符号a sub nThe nnth term of an arithmetic sequence
dd参数differenceThe common difference of an arithmetic sequence
SnS_n求和符号S sub nThe sum of the first nn terms of an arithmetic sequence
m,n,p,qm, n, p, q变量-Positive integers denoting term numbers

中英对照

中文术语英文术语音标说明
等差中项arithmetic mean/ˌærɪθˈmetɪk miːn/The arithmetic average of two numbers
对称性symmetry/ˈsɪmətri/The symmetric relationship between terms of a sequence
子数列subsequence/ˈsʌbsiːkwəns/A sequence extracted from the original one by a rule
二次函数quadratic function/kwɒˈdrætɪk ˈfʌŋkʃən/A function of the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c
最值extreme value/ɪkˈstriːm ˈvæljuː/The maximum or minimum value