CHAPTER 10
LESSON 1
Date ___________________
Review of Arithmetic Sequences and Series
AW Principles of Math 10: 1.2, 1.3
MP 6.2
Objective:
 To revisit arithmetic sequences and series from Principles of Math 10
Example 1:
Consider the sequence 3, 7, 11, 15, 19, …
What is the 101st term of the sequence?
The key feature of an arithmetic sequence is that there is a common difference d between
any two consecutive terms. To obtain any term, add d to the preceding term or subract d
from the following term.
th
Recall the formula for the n term ( tn ) of an arithmetic sequence:
tn  a  (n 1)d
where a is the first term ( t1 ), and d is the common difference between terms.
In this case, then, we have a  3, d  4, and n  101.
Example 2:
How many terms are there in the arithmetic sequence 4, 15, 26, … , 2853 ?
We will use the formula tn  a  (n 1)d and will solve for the variable n.
Example 3:
The 3rd term of an arithmetic progression is 13, and the 500th term is 2498. Find the first
two terms.
This results in a system of two equations in two unknowns:
Subtracting equation (1) from equation (2) we can eliminate the variable a and find d.
Thus, from equation (1), we can solve for a.
Example 4:
Find the sum of 101 terms of the series 3  7 1115 19  ... .
Recall the formula for the sum of n terms of an arithmetic series.
n
S  2 2a  (n  1)d
Example 5:
Find the sum of the arithmetic series S  2  5  8 11 ... 149 .
This time, we will use the following variation of the arithmetic series sum formula.
n
S  2 (a  t n )
We know that a  2 and tn 149. However, we will need to find the value for n (i.e., the
number of terms in the series) using the formula tn  a  (n 1)d .
Now we substitute this value for n into the above sum formula.