Example

Example: Finding the First Term and Common Difference Given Two Terms

Find the first term and common difference of an arithmetic sequence whose fifth term is 19 and whose eleventh term is 37, and give the formula for the general term.

Because two terms are known but neither a1a_1 nor dd is given, substitute each piece of information into the general term formula an=a1+(n1)da_n = a_1 + (n - 1)d to build a system of two equations.

Write the equations using n=5n = 5 and n=11n = 11:

19=a1+(51)d19 = a_1 + (5 - 1)d

37=a1+(111)d37 = a_1 + (11 - 1)d

Simplify:

19=a1+4d19 = a_1 + 4d

37=a1+10d37 = a_1 + 10d

To eliminate a1a_1, multiply the first equation by 1-1 and add the two equations:

19=a14d-19 = -a_1 - 4d

37=a1+10d37 = a_1 + 10d

Adding gives 18=6d18 = 6d, so d=3d = 3.

Substitute d=3d = 3 back into 19=a1+4d19 = a_1 + 4d:

19=a1+4(3)19 = a_1 + 4(3)

19=a1+1219 = a_1 + 12

a1=7a_1 = 7

Now write the general term using a1=7a_1 = 7 and d=3d = 3:

an=7+(n1)(3)a_n = 7 + (n - 1)(3)

an=7+3n3a_n = 7 + 3n - 3

an=3n+4a_n = 3n + 4

The first term is a1=7a_1 = 7 and the common difference is d=3d = 3. The general term of the sequence is an=3n+4a_n = 3n + 4.

This example demonstrates how knowing any two terms of an arithmetic sequence — even when neither a1a_1 nor dd is given directly — is enough to recover both values by setting up and solving a system of linear equations derived from the general term formula.

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Updated 2026-05-26

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Ch.12 Sequences, Series and Binomial Theorem - Intermediate Algebra @ OpenStax

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