The concept of arithmetic progression is used to calculate numbers between intervals. There are 1800000 numbers divisible by 5 between 1000000 and 9999999
The question is an illustration of an arithmetic progression with the following parameters.
[tex]a = 1000000[/tex] ---- the first term
[tex]L = 9999999[/tex] --- the last term
[tex]d = 5[/tex] -- the common difference
So, we are to solve for the number of terms; n.
The last term of an arithmetic progression is:
[tex]L = a + (n - 1)d[/tex]
Substitute known values
[tex]9999999 = 1000000 + (n - 1) \times 5[/tex]
Collect like terms
[tex]9999999 - 1000000 = (n - 1) \times 5[/tex]
[tex]8999999 = (n - 1) \times 5[/tex]
Divide both sides by 5
[tex]1799999.8 = n - 1[/tex]
Add 1 to both sides
[tex]1 + 1799999.8 = n[/tex]
[tex]1800000.8 = n[/tex]
Remove the decimal (do not approximate)
[tex]n = 1800000[/tex]
Hence, there are 1800000 numbers divisible by 5 between 1000000 and 9999999
Read more about progression at:
https://brainly.com/question/3927222
