Arithmetic Progression, number of terms, sum of terms etc

NO HIGHLIGHT: INCORRECT
HIGHLIGHT: CORRECT

Hi dave13 , a generous share here!
I'm responding to your question from this page.

You asked whether your formula for finding number of terms were correct.

The good news: You caught a discrepancy.

The oops news: No, in one part, your formula for finding number of terms is not correct

The great news: your formula is correct in another part

Here is the INCORRECT formula:

The correct formula is \(\frac{LastTerm-FirstTerm}{increment}+ 1\)

With small numbers, the formula's logic is apparent.

How many integers are there from 0 to 11, inclusive?
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
There are 12 integers

Your formula = 6.5 whoops

\(\frac{LastTerm-FirstTerm}{increment}+ 1\)

\(\frac{11-0}{1} + 1= (11 + 1) = 12\)

The foundation of the rule is evident: subtraction does not tell us the whole story.

When we subtract, we really find one fewer than the actual number of terms between two numbers. We're not including one endpoint. So we add one.

Increments: Not needed for consecutive integers

After (High - Low), with consecutive integers ONLY, there is no need to divide by the increment of 1. Any number divided by 1 is that number.

Increments: even and odd
Even and odd integers have a common difference, an increment, of 2.

For multiples, including even and odd integers, division by increment is necessary.

How many odd terms between 3 and 12, inclusive?
3, 5, 7, 9, 11
There are 5 such terms

In this case, we "find" the Last Term (last integer before 12 that is odd). With odds/ evens, finding first and last terms is easy.

Formula for \(n\) odd integers between 3 and 12?

\((\frac{LastTerm-FirstTerm}{increment}+ 1)\)
\(=(\frac{11-3}{2}+1)=(\frac{8}{2}+1)=(4+1)=5\)

You cannot use 12. It's not the Last Term.

Number of even integers between 45 and 699?
First Term: 46
Last Term: 698
\(\frac{698-46}{2}=\frac{652}{2}=326\) + 1 = 327

Increment for multiples
(Same as yours except for how to find Last Term, or "High")

Number of multiples of 7 between 12 and 1720, inclusive?Well, we can't list them.

Increment: 7 (common difference)
First Term: 14
Last Term: ???

To find the last multiple of 7 before 1720:
Divide 1720 by 7, to one decimal point
1720/7 = 245.7
We need the integer part: just multiply 7 by 245.
That yields the last multiple of 7 < 1720
(7*245)= 1,715
Last Term: 1,715

\(\frac{1715-14}{7}=\frac{1,701}{7}=243\) + 1 = 244
multiples of 7 between 12 and 1720.

Hope that answers your question. +1 for being generous with your work!