Arithmetic Progression and Sums

Sequence

A Sequence is a set of things (usually numbers) that are in order.

Arithmetic Progression

In an Arithmetic Progression the difference between one term and the next is a constant.
In other words, we just add the same value each time ... infinitely.

Example:

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number.

In General we could write an arithmetic sequence like this:

{a, a+d, a+2d, a+3d, ... }

where:

• a is the first term, and
• d is the difference between the terms (called the "common difference")

Example: (continued)

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

Has:

• a = 1 (the first term)
• d = 3 (the "common difference" between terms)

And we get:

{a, a+d, a+2d, a+3d, ... }

{1, 1+3, 1+2×3, 1+3×3, ... }

{1, 4, 7, 10, ... }

Rule

We can write an Arithmetic Sequence as a rule:

T_n = a + (n-1)d

Example 1: Calculate the 4th term for

3, 8, 13, 18, 23, 28, 33, 38, ...

This sequence has a difference of 5 between each number.

The values of a and d are:

• a = 3 
• d = 5

So, the 4th term can be calculated:
 

T_n = a + (n-1)d

T4 = a + (4-1)d 

     = a + 3d

     = 3 + 3(5)

     =18

 

 

Example 2: If T₃ = 8 and T₆ = 17, find T₁₄.

Solution:

T_{3 = a + 2d}

   _{8 = a + 2d -------①}

T_{6  = a +5d} 

_{17 = a + 5d --------②}

 

_{② - ①}

_{17-8  = a-a  + (5d-2d)}

_{9 = 3d}

 _d=3

_{substitute d=3 into ①}

_{8 = a + 2(3)}

_{a = 2}

 

T_{14  = a +13d}

         _{= 2 + 13(3)}

_{T14  = 41}

 

Summing an Arithmetic Series

To sum up the terms of this arithmetic sequence:

a + (a+d) + (a+2d) + (a+3d) + ...

use this formula:

  

Example1: 

Add up the first 10 terms of the arithmetic sequence:

{ 1, 4, 7, 10, 13, ... }

The values of a, d and n are:

• a = 1
• d = 3
• n = 10 (how many terms to add up)
• S10 = 10/2 [2(1) + ( 10-1) 3]
•        =  5 [ 2+9(3)]
•        = 5[29]
•        = 145

Example2:

0.6, 1.7, 2.8, ……………… to 100 terms

Solution:
a = 0.6, d = 1.1 and n = 100
Sum of n terms can be given as follows:

Thus, sum of the 100 term,
S₁₀₀= 5505

Problem 1:
The first term of an arithmetic sequence is equal to 200 and the common difference is equal to
-10. Find the value of the 20 th term

 

Solution to Problem 1:

Use the value of the common difference d = -10 and the first term a₁ = 200 in the formula for the n th term given above and then apply it to the 20 th term

T₂₀ = 200 +(20 - 1 ) (-10) 

      = 10

   

Problem 2:
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.

 

Solution to Problem 2:

• We use the n th term formula for the 6 th term, which is known, to write
  T₆ = 52    
•     = a +(6 - 1 ) (10)
• 52 = a +50
• a   = 2
  
  
• The above equation allows us to calculate a.
  
  a = 2
  
  
• Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows.
  
  T₁₅ = 2 + (15 - 1)(10) = 142

Problem 3:
Find the sum of all the integers from 1 to 1000.

 

Solution to Problem 3:

• The sequence of integers starting from 1 to 1000 is given by
  
  1 , 2 , 3 , 4 , ... , 1000
  
  
• The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above).
  
  s₁₀₀₀ = 1000 (1 + 1000) / 2 = 500500

Problem 4:

Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

 

Solution to Problem 4:

• The first few terms of a sequence of positive integers divisible by 5 is given by
  
  5 , 10 , 15 , ...
  
  
• The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows
  
  1555 = a₁ + (n - 1 )d
  
  
• Substitute a₁ and d by their values
  
  1555 = 5 + 5(n - 1 )
  
  
• Solve for n to obtain
  
  n = 311
  
  
• We now know that 1555 is the 311 th term, we can use the formula for the sum as follows
  
  s₃₁₁ = 311 (5 + 1555) / 2 = 242580 

We can use this fomula to calculate the sum of the AP :

 where a = first term,  l = last term

 Problem 5:

Find the sum of the first 50 even positive integers. 

Solution to Problem 5:

• The sequence of the first 50 even positive integers is given by
  
  2 , 4 , 6 , ...
  
  
• The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term
  T₅₀ = 2 +  49(2) = 100
  
  
• We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms
  
  s₅₀ = 50 (2 + 100) / 2 = 2550