- Arithmetic Progressions

A sequence \(a_1, a_2, a_3, …, a_{n-1}, a_n, …\) is called an arithmetic sequence, or arithmetic progression, if there exists a constant \(d\), called the common difference, such that

\[a_n-a_{n-1}=d\]
That is
\(a_1\)
\(a_2=a_1+d\)
\(a_3=a_2+d\)
\(=a_1+d+d\)
\(=a_1+2d\)
Therefore,
\(a_n= a_1+(n-1)d\) for every \(n>1\)

LESSON 1
Find the common difference for each of the following arithmetic progressions.
(a) 3, 5, 7, 9, 11, …
(b) 8, 3, -2, -7, …
(c) \(2b, 5b, 8b, 11b, …\)
SOLUTION
(a) \(a_1=3, a_2=5, a_3=7,…\)
\(d=a_2-a_1=5-3=2\)

(b) \(a_1=8, a_2=3, a_3=-2, a_4=-5\)
\(d=a_3-a_2=-2-3=-5\)

(c) \(a_1=2b, a_2=5b, a_3=8b, a_4=11b, …\)
\(d=a_4-a_3=11b-8b=3b\)
NB: Any pair of consecutive terms can be used.

LESSON 2
Prove that the sequence 3, 7, 11, 15, … is an arithmetic progression.
SOLUTION
We need to show that \(a_n-a_{n-1}\) is a constant.
\(a_n=4n-1\)
\(a_{n-1}=4(n-1)-1\)
\(=4n-5\)
\(a_n-a_{n-1}=(4n-1)-(4n-5)\)
\(=4\)
Therefore, \(d=4\)

LESSON 3
The sum, \(S_n\), of the first n terms of a sequence is given by \(S_n=n(5n-2)\). Show that the sequence is an arithmetic progression with common difference 10.
SOLUTION
  • Determine an expression for \(a_n\).
\(a_n=S_n-S_{n-1}\)
\(=n(5n-2)-[(n-1)(5(n-1)-2)]\)
\(=5n^2-2n-[(n-1)(5n-5-2)]\)
\(=5n^2-2n-[(n-1)(5n-7)]\)
\(=5n^2-2n-(5n^2-7n-5n+7)\)
\(=5n^2-2n-(5n^2-12n+7)\)
\(=5n^2-2n-5n^2+12n-7\)
\(=10n-7\)
  • Use \(a_n\) to write an expression for \(a_{n-1}\)
\(a_{n-1}=10(n-1)-7\)
\(=10n-10-7\)
\(=10n-17\)
  • Determine the common difference, \(d\).
\(d=a_n-a_{n-1}\)
\(=(10n-7)-(10n-17)\)
\(=10\)

LESSON 4
If the first three terms of an arithmetic progression are 5, 9, and 13, what is the value of the 10th term?
SOLUTION
\(a_1=5, a_2=9, a_3=13\)
The common difference, \(d\), is 4
\(a_{10}=a_1+(n-1)d\)
\(=5+(10-1)(4)\)
\(=41\)

SUM FORMULAE FOR FINITE ARITHMETIC SEQUENCE
If \(a_1, a_2, a_3,…, a_n\) is a finite arithmetic sequence, then the corresponding series
\(a_1+a_2+a_3+⋯+a_n\) is called a finite arithmetic series. The sum of the first \(n\) terms of the series, which we denote \(S_n\), would be stated as
\(\displaystyle S_n={n\over2}[2a_1+(n-1)d]\)

LESSON 1
Find the sum of the even numbers from 50 to 120 inclusive.
SOLUTION
\(a_1=50\) This is the first even number
\(d=2\) Even numbers differ by 2
\(a_n=120\) This is the last even number
  • Determine the value of \(n\).
\(a_1+(n-1)d=120\)
\(50+2(n-1)=120\)
\(2(n-1)=70\)
\(n-1=35\)
\(n=36\)
  • Substitute required values into
\(\displaystyle S_n={n\over2} [2a_1+(n-1)d]\)
\(\displaystyle S_{36}={36\over2} [2(50)+(36-1)(2)]\)
\(S_{36}=3060\)

LESSON 2
The last term of an arithmetic progression of 20 terms is 295 and the common difference is 4. Calculate the sum of the progression.
SOLUTION
\(n=20, a_{20}=295, d=4\)
  • We need to determine \(a_1\)
\(a_20=a_1+(n-1)d\)
\(a_1+(20-1)(4)=295\)
\(a_1=219\)
  • Substitute required values into
\(S_n={n\over 2} [2a_1+(n-1)d]\)
\(S_{20}={20\over2} [2(219)+(20-1)(4)]\)
\(S_{20}=5140\)

LESSON 3
The sum of the first 6 terms of an arithmetic progression is 54.75 and the sum of the next 6 terms is 63.75. Find the common difference and the first term.
SOLUTION
Use the information given about \(S_6\) to determine an equation in \(a_1\) and \(d\)
\(S_6=54.75\)
\(\displaystyle {6\over2} [2a_1+(6-1)d]=54.75\)
\(6a_1+15d=54.75 \:\:\;\:\:(1)\)
  • Use the information given about \(S_{12}\) to determine an equation in \(a_1\) and \(d\)
\(S_{12}=54.75+63.75=118.5\)
\(\displaystyle {12\over2} [2a_1+(12-1)d]=118.5\)
\(12a_1+66d=118.5 \:\:\:\:\:(2)\)

  • Solve (1) and (2) simultaneously
\(6a_1+15d=54.75\)
\(12a_1+66d=118.5\)

\((1) \times 2\):
\(12a_1+30d=109.5\)
\(12a_1+66d=118.5\)
\(-36d=-9\)
\(\displaystyle d={1\over4}\)
\(a=8.5\)