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Pre-Calculus Sequences and series (advanced)

Arithmetic series: nth term and partial sums

20 practice questions 0 video lessons Theory + worked examples

Arithmetic Series

California Pre-Calculus • Standard A-SSE.4 • Sequences & Series

Arithmetic Series is a topic in Sequences & Series in the California Common Core State Standards. It is aligned to Standard A-SSE.4, which requires students to derive and use the formula for the sum of an arithmetic series.

An arithmetic series sums a sequence with a constant difference, using \(a_n=a_1+(n-1)d\) and \(S_n=\dfrac{n}{2}(a_1+a_n)\).

California Pre-Calculus › Sequences & Series › Arithmetic Series  —  Standard A-SSE.4

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Theory

An arithmetic sequence has a constant common difference \(d\) between consecutive terms. Its \(n\)th term is

\[a_n=a_1+(n-1)d.\]

The sum of the first \(n\) terms, an arithmetic series, is

\[S_n=\dfrac{n}{2}(a_1+a_n)=\dfrac{n}{2}\big(2a_1+(n-1)d\big).\]
The sum formula is average \(\times\) count: \(\dfrac{a_1+a_n}{2}\) is the average term, times \(n\) terms.
Arithmetic sequence: constant difference An arithmetic sequence has a constant difference d between consecutive terms. +d +d +d +d +d
Equal steps: a constant difference \(d\).
Arithmetic formulas Arithmetic formulas Arithmetic formulas aₙ = a₁ + (n−1)d Sₙ = n2 (a₁ + aₙ) = n2 (2a₁ + (n−1)d)
The arithmetic term and sum formulas.

The nth term and the partial sum:

\[a_n=a_1+(n-1)d,\qquad S_n=\dfrac{n}{2}(a_1+a_n)\]
nth term is a1 plus n minus 1 times d; the sum is n over 2 times a1 plus an
Use the second sum form \(\dfrac{n}{2}(2a_1+(n-1)d)\) when you don't yet know \(a_n\).

How to work with arithmetic series

  1. Find \(d\) as the difference of consecutive terms.
  2. nth term: \(a_n=a_1+(n-1)d\).
  3. Sum: \(\dfrac{n}{2}(a_1+a_n)\), or the expanded form if \(a_n\) is unknown.
  4. Solve for \(n\) from a term or a sum when required.
Example 1 — The nth term
Find the 20th term of \(3,7,11,\dots\).
Solution

\(a_1=3,\ d=4\); use \(a_n=a_1+(n-1)d\).

a_{20}\(=\)3+(20-1)4
\(=\)3+76=79
the 20th term is 79
Example 2 — Partial sum
Find the sum of the first 20 terms of \(3,7,11,\dots\).
Solution

Use \(S_n=\dfrac{n}{2}(a_1+a_n)\) with \(a_{20}=79\).

S_{20}\(=\)\dfrac{20}{2}(3+79)
\(=\)10\cdot 82=820
the sum of the first 20 terms is 820
Example 3 — Sum without the last term
Sum the first 15 terms of an arithmetic sequence with \(a_1=5,\ d=2\).
Solution

Use \(S_n=\dfrac{n}{2}(2a_1+(n-1)d)\).

S_{15}\(=\)\dfrac{15}{2}(2\cdot 5+14\cdot 2)
\(=\)\dfrac{15}{2}(38)=285
the sum is 285
Example 4 — Find the number of terms
How many terms of \(2,5,8,\dots\) are needed to reach a last term of \(35\)?
Solution

Solve \(a_n=35\) with \(a_1=2,\ d=3\).

2+(n-1)3\(=\)35
(n-1)3\(=\)33
n\(=\)12
12 terms are needed

Common pitfalls

\((n-1)d\), not \(nd\). The first term already counts as term 1.
Constant difference, not ratio. Arithmetic adds; geometric multiplies.
Match the sum form to what you know — use \(2a_1+(n-1)d\) when \(a_n\) is unknown.

Frequently asked questions

What is an arithmetic sequence?

A sequence with a constant difference \(d\) between consecutive terms; the \(n\)th term is \(a_1+(n-1)d\).

How do you find the sum of an arithmetic series?

\(S_n=\dfrac{n}{2}(a_1+a_n)\) — the average of the first and last term times the number of terms.

What is the common difference?

The fixed amount \(d\) added to each term to get the next.

Why is it (n - 1)d and not nd?

Because the first term needs no addition; only \(n-1\) steps of \(d\) reach the \(n\)th term.