Progressions: (lesson 1 of 2)
Arithmetic Progressions
Definition:
By an arithmetic progression of m terms, we mean a finite sequence of the form
a,a+d,a+2d,a+3d,...,a+(m−1)d.
The real number a is called the first term of the arithmetic progression, and the real number d is called
the difference of the arithmetic progression.
Example 1:
Consider the sequence of numbers
1,3,5,7,9,11,13,15,17,19,21,23
The property of this sequence is that the difference between successive terms is constant and equal to 2.
Here we have: a=1; d=2.
Example 2:
Consider the sequence of numbers:
2,5,8,11,14,17,20,23,26,29,32
The property of this sequence is that the difference between successive terms is constant and equal to 3.
Here we have: a=2; d=3.
General term of arithmetic progression:
The general term of an arithmetic progression with first term a1 and common difference d is:
ak=a1+(k−1)d
Example 3: Find the general term for the arithmetic sequence
−1,3,7,11,... and then find a12.
Solution:
Here a1=1. To find d subtract any two adjacent terms: d=7−3=4. The general term is:
akakakak=a1+(k−1)d=−1+(k−1)⋅4=−1+4k−4=4k−5
To find a12, let k=12.
a12=4⋅12−5=43
Example 4: If a3=8 and a6=17, find a14.
Solution:
Use the formula for ak with the given terms
a38a617=a1+(3−1)⋅d=a1+2d=a1+(6−1)⋅d=a1+5d
This gives us a system of two equations with two variables. By solving them, we can find that a1=2 and d=3.
Use the formula for ak to find a14
aka14a14=a1+(k−1)⋅d=2+(14−1)⋅3=41
Exercise:
Sum of an arithmetic progression:
The sum of the n terms of an arithmetic progression with first term a1 and common difference d is:
an=2n[2a1+(n−1)d]
Also, the sum of an arithmetic progression is equal to
Sn=2n(a1+an)
Example 5:
Find the sum of the 10 terms of the arithmetic progression if a1=5 and d=4.
Solution:
SnS10S10S10=2n[2a1+(n−1)d]=210[2⋅5+(10−1)⋅4]=5⋅46=230
Example 6: Find 1+2+3+...+100
Solution:
In this example, we have: a1=1, d=1, n=100, a100=100. The sum is:
SnS100S100=2n(a1+an)=2100(1+100)=5050
Exercise: