General Term Formula of Arithmetic Sequences

The general term formula is the core tool of arithmetic sequences. It allows us to directly find any term of the sequence without computing term by term.

Derivation of the Formula

For an arithmetic sequence {an}\{a_n\}, let the first term be a1a_1 and the common difference be dd. Let us derive the expression for the nnth term ana_n.

According to the definition of an arithmetic sequence:

a2=a1+da3=a2+d=a1+2da4=a3+d=a1+3da5=a4+d=a1+4d⋮\begin{aligned} a_2 &= a_1 + d \\ a_3 &= a_2 + d = a_1 + 2d \\ a_4 &= a_3 + d = a_1 + 3d \\ a_5 &= a_4 + d = a_1 + 4d \\ &\vdots \end{aligned}

Observe the pattern: from a1a_1 to ana_n, we need to add (n−1)(n-1) common differences dd.

General term formula of an arithmetic sequence
an=a1+(n−1)da_n = a_1 + (n-1)d

This is a great question! From a1a_1 to ana_n, we need to go through (n−1)(n-1) steps, not nn steps.

Imagine climbing stairs: from the 1st floor to the nnth floor, you need to climb (n−1)(n-1) steps. For example, from the 1st floor to the 3rd floor, you only climb 2 steps (1→2, 2→3), not 3.

Similarly, from a1a_1 to ana_n, we need to add the common difference (n−1)(n-1) times:

  • a1a_1 to a2a_2: add 1 time
  • a1a_1 to a3a_3: add 2 times
  • a1a_1 to ana_n: add (n−1)(n-1) times

Variations of the General Term Formula

The general term formula can also be written in other forms:

Form 1: Based on an Arbitrary Term

If the mmth term ama_m is known, then:

an=am+(n−m)da_n = a_m + (n-m)d

This formula is especially useful when a certain term (not necessarily the first) is known.

Form 2: The Linear Function Form

Expanding the general term formula:

an=a1+(n−1)d=dn+(a1−d)a_n = a_1 + (n-1)d = dn + (a_1 - d)

Let k=dk = d and b=a1−db = a_1 - d. Then:

an=kn+ba_n = kn + b

This shows that the general term formula of an arithmetic sequence is a linear function of nn (when d≠0d \neq 0).

Worked Examples

Example 1: Given the first term and common difference, find a certain term

In the arithmetic sequence {an}\{a_n\}, the first term is a1=3a_1 = 3 and the common difference is d=2d = 2. Find the 10th term a10a_{10}.

Solution:

a10=a1+(10−1)d=3+9×2=3+18=21a_{10} = a_1 + (10-1)d = 3 + 9 \times 2 = 3 + 18 = 21

Example 2: Given two terms, find the general term formula

In the arithmetic sequence {an}\{a_n\}, a3=7a_3 = 7 and a7=15a_7 = 15. Find the general term formula.

Solution:

Using the general term formula:

{a3=a1+2d=7a7=a1+6d=15\begin{cases} a_3 = a_1 + 2d = 7 \\ a_7 = a_1 + 6d = 15 \end{cases}

Subtracting the two equations: 4d=8⇒d=24d = 8 \quad \Rightarrow \quad d = 2

Substituting into the first equation: a1+2×2=7⇒a1=3a_1 + 2 \times 2 = 7 \quad \Rightarrow \quad a_1 = 3

Therefore, the general term formula is: an=3+(n−1)×2=2n+1a_n = 3 + (n-1) \times 2 = 2n + 1

Example 3: Determine whether a number is a term of the sequence

In the arithmetic sequence {an}\{a_n\}, a1=5a_1 = 5 and d=3d = 3. Is 101101 a term of this sequence? If so, which term?

Solution:

Suppose 101101 is the nnth term. Then: an=5+(n−1)×3=101a_n = 5 + (n-1) \times 3 = 101 3n+2=1013n + 2 = 101 3n=993n = 99 n=33n = 33

Since n=33n = 33 is a positive integer, 101101 is the 33rd term of this sequence.

Practice Problems

Exercise 1

In the arithmetic sequence {an}\{a_n\}, a1=−5a_1 = -5 and d=3d = 3. Find a20a_{20}.

arithmetic sequencegeneral term formula

Approach: Use the general term formula an=a1+(n−1)da_n = a_1 + (n-1)d directly.

Steps:

a20=a1+(20−1)d=−5+19×3=−5+57=52a_{20} = a_1 + (20-1)d = -5 + 19 \times 3 = -5 + 57 = 52

Answer: a20=52a_{20} = 52

Exercise 2

In the arithmetic sequence {an}\{a_n\}, a4=10a_4 = 10 and a9=25a_9 = 25. Find the first term a1a_1 and the common difference dd.

arithmetic sequencegeneral term formula

Approach: Set up a system of equations and solve.

Steps:

According to the general term formula:

{a4=a1+3d=10a9=a1+8d=25\begin{cases} a_4 = a_1 + 3d = 10 \\ a_9 = a_1 + 8d = 25 \end{cases}

Subtracting the two equations: 5d=15⇒d=35d = 15 \quad \Rightarrow \quad d = 3

Substituting into the first equation: a1+3×3=10⇒a1=1a_1 + 3 \times 3 = 10 \quad \Rightarrow \quad a_1 = 1

Answer: a1=1a_1 = 1, d=3d = 3

Exercise 3

In the arithmetic sequence {an}\{a_n\}, a1=2a_1 = 2 and d=5d = 5. Is 9797 a term of this sequence?

arithmetic sequencegeneral term formula

Approach: Suppose 9797 is the nnth term and check whether nn is a positive integer.

Steps:

Suppose an=97a_n = 97. Then: 2+(n−1)×5=972 + (n-1) \times 5 = 97 5n−3=975n - 3 = 97 5n=1005n = 100 n=20n = 20

Since n=20n = 20 is a positive integer, 9797 is the 20th term of this sequence.

Answer: Yes, 9797 is the 20th term

Exercise 4

The arithmetic sequence {an}\{a_n\} satisfies a2+a4=14a_2 + a_4 = 14 and a3+a6=21a_3 + a_6 = 21. Find a1a_1 and dd.

arithmetic sequencegeneral term formulasum of first n terms

Approach: Use the general term formula to set up a system of equations.

Steps:

According to the general term formula:

a2+a4=(a1+d)+(a1+3d)=2a1+4d=14a3+a6=(a1+2d)+(a1+5d)=2a1+7d=21\begin{aligned} a_2 + a_4 &= (a_1 + d) + (a_1 + 3d) = 2a_1 + 4d = 14 \\ a_3 + a_6 &= (a_1 + 2d) + (a_1 + 5d) = 2a_1 + 7d = 21 \end{aligned}

Simplifying:

{2a1+4d=14⇒a1+2d=72a1+7d=21\begin{cases} 2a_1 + 4d = 14 \quad \Rightarrow \quad a_1 + 2d = 7 \\ 2a_1 + 7d = 21 \end{cases}

Subtracting the two equations: 3d=7⇒d=733d = 7 \quad \Rightarrow \quad d = \frac{7}{3}

Substituting into the first equation: 2a1+4×73=142a_1 + 4 \times \frac{7}{3} = 14 2a1=14−283=42−283=1432a_1 = 14 - \frac{28}{3} = \frac{42-28}{3} = \frac{14}{3} a1=73a_1 = \frac{7}{3}

Answer: a1=73a_1 = \frac{7}{3}, d=73d = \frac{7}{3}


Summary

Symbols Used in This Article

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
a1a_1Element symbola sub 1The first term of an arithmetic sequence
ana_nElement symbola sub nThe nnth term of an arithmetic sequence
ddParameterdifferenceThe common difference of an arithmetic sequence
nnVariablenThe term number, a positive integer

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