## - 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
$$6a_1+15d=54.75$$
$$12a_1+66d=118.5$$

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