Arithmetic Progression - equal value increase or decrease of terms

In the previous lessons, we have looked at various problems involving Sequence and Series. We have also seen some worked examples of both. Now its time to go up a little bit an introduce a type of Sequence called arithmetic Progression. Lets get into some details then.

Arithmetic Progression (often shortened A.P) is a type of sequence easily identified by the pattern of the next terms in the set. A discreet value is usually added or subtracted to obtain the next term of the sequence. That value could be a whole number, fraction or some other quantities. That is the main identifying feature of arithmetic progressions. This number that is added or removed to get the next term is called the common difference (d). Below are examples of Arithmetic Progressions:

• 5,15,25,35,.... (The common difference here is +10 or -10 depending on the direction you followed)

• 4,8,14,16,.... (The common difference here is +4 or -4 depending on the direction you follow)

• 7,14,21,28,.... (The common difference here is +7 or -7 depending on the direction you follow)

Solving for the nth of an Arithmetic Progression

First of all, you need to understand the important components of an Arithmetic progression and their symbols. So here they are below:

• First term: The first term of an arithmetic progression is represented by the letter a
• Common difference: The common difference is denoted by the letter d
• number of terms: The letter n is used to denote the number of terms.
• Terms of an AP: The terms of any Arithmetic progression is represented by the letter T

Having understood these important elements, below is the derived formula used to find the Terms of an Arithmetic progression:

We will now use this formula to solve some specific examples:

Example 1: In the following arithmetic progression, find the 25th term.
8,10,12,14,16.

Solution:
T_n = a + (n - 1)d
a = 8
n=25
d=16-14 or12-10 0r 10-2 = 2


T₂₅ = 8 + (25 - 1)2
Expanding the bracket
T₂₅ = 8 + 24 x 2
T₂₅ = 8 + 48
T₂₅ = 56

Example 2: In the following A.P 30,24,18,12... solve for the 15th term.
Solution:
T_n = a + (n - 1)d
a = 30
n=15
d=24-30 or 18-24 = -6


T₁₅ = 30 + (15 - 1)-6
Expanding the bracket
T₁₅ = 30 + 14 x -6
T₁₅ = 30 + -84
T₁₅ = 30 - 84
T₁₅ = 30 - 84
T₁₅ = -54

Example 3: Solve for the 20th term of of the following AP -12, -8,-4,0
Solution:
T_n = a + (n - 1)d
a = -12
n=20
d=-8-(-12) or -4-(-8) = 4


T₂₀ = -12 + (20 - 1)4
Expanding the bracket
T₂₀ = -12 + 19 x 4
T₂₀ = -12 + 19 x 4
T₂₀ = -12 + 76
T₂₀ = 64

Conclusion

We have seen in this introductory course what arithmetic progression is and the important components - the first term, common difference and nth term. The three examples above also demonstrates how we can solve for any number of term using the derived formula. In the next presentation, we will try to look at the formula for finding the sum of terms of an arithmetic progression. We will also look as some worked examples too. Stay tuned and watch this space.