Arithmetic Sequences

Sequence is a list of numbers typically with a pattern.

2, 4, 6, 8, …

The first term in a sequence is denoted as a1, the second term is a2, and so on up to the nth term an.

Each number in the list called a term.

a1, a2, a3, a4, …

Finite Sequence has a fixed number of terms. {2, 4, 6, 8}

A sequence that has infinitely many terms is called an infinite sequence.

{2, 4, 6, 8,…}

Algebraically, a sequence can be written as an explicit formula or as a recursive formula.

Explicit formulas show how to find a specific term number (n).

Recursive formula show how to get from a given term (an-1) to the next term (an)

An Arithmetic Sequence is a sequence where you use repeated addition (with same number) to get from one term to the next.

Ex: 4, 1, -2, -5, … is an arithmetic sequence

-3 -3 -3

The number that needs to be added each time to get to the next term is called the common difference

The common difference for the above arithmetic sequence is -3 .

Explicit formula for Arithmetic Sequence:

an = dn + (a1 – d)

Recursive formula for an Arithmetic Sequence:

a1 = #an = an-1 + d

First term

Common difference

Explicit FormulaSubstitute the values:an = -3n + (4 - -3)

So the explicit formula is: an = -3n + 7

The Recursive Formula is:

a1 = 4an = an-1 – 3

For the example: 4, 1, -2, -5, …

A series is the sum of ALL the terms of a sequence. (can be finite or infinite)

A partial sum is the sum of the first n terms of a

series…denoted Sn

𝑆𝑛=𝑛 (𝑎1+𝑎𝑛)

2

Number of termsFirst term Last term

How do you add these sequences of numbers?

For the example: 4, 1, -2, -5, …

1) Find S4.

2) Find S20. (If you don’t know the last term, use the explicit formula to find it!)

𝑺𝟒=𝟒(𝟒+−𝟓)

𝟐=−𝟐

𝑺𝟐𝟎=𝟐𝟎 (𝟒+−𝟓𝟑)

𝟐=−𝟒𝟗𝟎

Explicit formula is an = -3n + 7, so the 20th term is -3 (20) + 7 = -53

Example: For the arithmetic sequence 2, 6, 10, 14, 18, …

a) Write the explicit formula for the sequence.

b) Write the recursive formula for the sequence.

c) Find the 15th partial sum of the sequence (S15).

an = dn + (a1 – d)a1 = #an = an-1 + d

𝑆𝑛=𝑛 (𝑎1+𝑎𝑛)

2