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Greetings friends 
l
i decided to create a list of all formulas regarding arithmetic progression, number of terms, sums etc all in one post. i may have some questions and or inaccuaracies, so you are welcome to correct me or just say hi
1. HOW TO FIND NUMBER OF TERMS
\(x_n = a+d (n-1)\)
\(n\) = # of term
\(a\) = first term
\(d\) = distance
Example:
a= 3
d = 5
Question: find the 9th term
\(x_9 = 3+5 (n-1)\)
\(x_9 = 3+5 (n-1)\)
\(x_9 = 3+5n-5\)
\(x_9 = 5n-2\)
now plug in 9 into 5n-2
\(x_9 = 5*9-2 = 43\)
hence \(9th\) term is \(43\)
2. HOW TO FIND THE SUM OF TERMS
SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)
Example
\(a\) = 1 the first term
\(d\) = 3 distance
\(n\) = 10 how many terms to add up
Question: what is the sum of 10 terms with distance 3 and the first term 1 ?
\(\frac{10}{2} (2 *1 +(10-1)3)\)
\(= 5(2+9·3) = 5(29) = 145\)
3.HOW TO FIND THE SUM OF THE FIRST CONSECUTIVE NUMBERS
\(\frac{n(n+1)}{2}\)
where \(n\) is number of terms
Example: what is the sum of the first 15 numbers ?
\(\frac{15(15+1)}{2}\) =\(238\)
4.HOW TO FIND THE SUM OF THE FIRST EVEN NUMBERS
\(n(n+1)\), where \(n\) is number of terms
2 + 4 + 5 = 12
n(n + 1) = 3(3 + 1) = 12
5. HOW TO FIND NUMBER OF TERMS FROM A TO Z
\(\frac{last..term - first..term}{2} +1\)
6. HOW TO FIND SUM OF ODD NUMBERS FROM A TO B
Step one: \(find..the..number...of..terms\)
Step two: \(\frac{first..term+last..term}{2}\) \(* number..of.. terms\)
7. NUMBER OF MULTIPLES X IN THE RANGE
\(\frac{last..multiple..of..x - first..multiple..of...x}{x}+1\)
Eg. how many multiples of 4 are there between 12 and 96?
\(\frac{96-12}{4}\)+1 = 22
8. Sum of the Squares of First N positive integers
\(n\) \(\frac{{n(n+1)(2n+1)}}{{6}}\)
Eg. Find the last digit of the number \(1^2+2^2+..........+99^2\)
Sum of the Squares of First \(99\) Positive Integers = \(\frac{{99(100)(198+1)}}{{6}}={33*50*199}\)
to be continued
by the way I myself have question
what is the difference between this \(\frac{n(n+1)}{2}\) and this formula SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\) ?
I will add more useful formulas later
li decided to create a list of all formulas regarding arithmetic progression, number of terms, sums etc all in one post. i may have some questions and or inaccuaracies, so you are welcome to correct me or just say hi
1. HOW TO FIND NUMBER OF TERMS
\(x_n = a+d (n-1)\)
\(n\) = # of term
\(a\) = first term
\(d\) = distance
Example:
a= 3
d = 5
Question: find the 9th term
\(x_9 = 3+5 (n-1)\)
\(x_9 = 3+5 (n-1)\)
\(x_9 = 3+5n-5\)
\(x_9 = 5n-2\)
now plug in 9 into 5n-2
\(x_9 = 5*9-2 = 43\)
hence \(9th\) term is \(43\)
2. HOW TO FIND THE SUM OF TERMS
SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)
Example
\(a\) = 1 the first term
\(d\) = 3 distance
\(n\) = 10 how many terms to add up
Question: what is the sum of 10 terms with distance 3 and the first term 1 ?
\(\frac{10}{2} (2 *1 +(10-1)3)\)
\(= 5(2+9·3) = 5(29) = 145\)
3.HOW TO FIND THE SUM OF THE FIRST CONSECUTIVE NUMBERS
\(\frac{n(n+1)}{2}\)
where \(n\) is number of terms
Example: what is the sum of the first 15 numbers ?
\(\frac{15(15+1)}{2}\) =\(238\)
4.HOW TO FIND THE SUM OF THE FIRST EVEN NUMBERS
\(n(n+1)\), where \(n\) is number of terms
2 + 4 + 5 = 12
n(n + 1) = 3(3 + 1) = 12
5. HOW TO FIND NUMBER OF TERMS FROM A TO Z
\(\frac{last..term - first..term}{2} +1\)
6. HOW TO FIND SUM OF ODD NUMBERS FROM A TO B
Step one: \(find..the..number...of..terms\)
Step two: \(\frac{first..term+last..term}{2}\) \(* number..of.. terms\)
7. NUMBER OF MULTIPLES X IN THE RANGE
\(\frac{last..multiple..of..x - first..multiple..of...x}{x}+1\)
Eg. how many multiples of 4 are there between 12 and 96?
\(\frac{96-12}{4}\)+1 = 22
8. Sum of the Squares of First N positive integers
\(n\) \(\frac{{n(n+1)(2n+1)}}{{6}}\)
Eg. Find the last digit of the number \(1^2+2^2+..........+99^2\)
Sum of the Squares of First \(99\) Positive Integers = \(\frac{{99(100)(198+1)}}{{6}}={33*50*199}\)
to be continued
by the way I myself have question
what is the difference between this \(\frac{n(n+1)}{2}\) and this formula SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\) ?
I will add more useful formulas later
27 Apr 2018, 17:40
Kudos
Bookmarks
dave13
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:
Quote:
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!
General Discussion
26 Mar 2018, 12:47
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dave13
Hello dave13,
Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)
Let's see,
SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)
plug in a = 1, d = 1
\(\frac{n}{2} (2*1+(n-1)1)\)
\(\frac{n}{2} (n+1)\)
\(n*(n+1) / 2\)
So we can say that SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\) is a generalized formula whose special case is
\(\frac{n(n+1)}{2}\) when we talk about first n consecutive integers.
Good initiative!
+1 kudos to you!
Best,
Gladi
