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Wednesday, February 18, 2015

Arithmetic Progression and Sums

Sequence


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


Sequence


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:

Tn = 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:
 
Tn = 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:

T3 = a + 2d
   8 = a + 2d -------①

T6  = 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

 

T14  = 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:
sum of n terms
10 arithmetic progression exercise solution
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
 

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