# Arithmetic Progression

During our elementary days, our teachers told us to count by 1’s(1,2,3..), by 2’s(2,4,6…), by 3’s(3,6,9..). Without our knowledge, we are creating a series called arithmetic progression. This series is easy to understand because if you are not sure you can enumerate the elements if you forgot the formula.

Nth term of arithmetic progression;

Given that the series has first term and common difference of r

nth term                                                formula

1st term                                                $a_1$

2nd term                                               $a_1+d$

3rd term                                                $a_1+2d$

4th term                                                $a_1+3d$

5th term                                                $a_1+4d$

Observe the pattern above about the nth term and the coefficient of r. We can tell that the coefficient of r is 1 less than the coefficient of the nth term. We can create a formula by looking at that pattern.

Formula:

Let $a_n$– nth term, $a_1$– 1st term, $d$– common difference

$a_n=a_1+(n-1)d$

Worked Problem 1:

Give the 2014th term of the series. 3, 5, 7…

Solution:

Based on our formula we need to know the first term and the common difference.

Given: $a_1=3$      $d=2$

$a_n=a_1+(n-1)d$ $a_{2014}=3+(2014-1)2$ $a_{2014}=4029$

Worked Problem 2:

The nth term of an arithmetic series is given by the formula $a_n=5+3n$. Find the following;

(a). first term

(b). 2014th term

(c). Common difference

Solution:

(a). first term:

$a_n=5+3n$

Let n=1

$a_1=5+3(1)$ $a_1=8$

(b)2014th term

$a_n=5+3n$

Let n=2014

$a_{2014}=5+3(2014)$ $a_{2014}=6047$

(c). Common difference

In $a_n=5+3n$ the common difference is the coefficient of n which is 3. If you want to check , try to solve for the second term and see for yourself. Technically, $a_n=5+3n$ came from the general formula $a_n=a_1+(n-1)d$ with known common difference d. Using distributive property we can rewrite the formula in $a_n=a_1+dn-d$. In the given expression $a_n=5+3n$, $5=a_1-d$ and $3n=dn$ making d=3.

Worked Problem 3:

The 4th term of an arithmetic progression is 18 and the 10th term is 42. Find the 100th term.

Solution:

$a_n=a_1+(n-1)d$ $a_{10}=a_{4}+(n-1)d$

Now, we count how many terms from 4th term to 10th term.(4,5,6,7,8,9,10) making n=7.

$a_{10}=a_{4}+(7-1)d$ $42=18+6d$ $d=4$

From 4th term to 100th terms there are 100-3=97 terms. That is the 100 terms minus the first three terms.

$a_n=a_1+(n-1)d$ $a_{100}=a_4+(97-1)4$ $a_{100}=18+(96)4$ $a_{100}=402$