# 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

1^{st} term

2^{nd} term

3^{rd} term

4^{th} term

5^{th} term

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 – nth term, – 1^{st} term, – common difference

**Worked Problem 1:**

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

**Solution:**

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

Given:

**Worked Problem 2:**

The nth term of an arithmetic series is given by the formula . Find the following;

(a). first term

(b). 2014^{th} term

(c). Common difference

**Solution:**

(a). first term:

Let n=1

(b)2014^{th} term

Let n=2014

(c). Common difference

In 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, came from the general formula with known common difference d. Using distributive property we can rewrite the formula in . In the given expression , and making d=3.

**Worked Problem 3:**

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

**Solution:**

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

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

### Dan

#### Latest posts by Dan (see all)

- 2016 MMC Schedule - November 4, 2015
- 2014 MTAP reviewer for Grade 3 - September 30, 2015
- 2015 MTAP reviewer for 4th year solution part 1 - August 22, 2015