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Arithmetic Progression, number of terms, sum of terms etc

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Arithmetic Progression, number of terms, sum of terms etc  [#permalink]

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New post Updated on: 24 Jul 2018, 11:04
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2
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

\(\frac{n(n+2)}{4}\)

where \(n\) is number of terms

\(\frac{15(15+2)}{4}\) = \(8\)



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 :)

Originally posted by dave13 on 24 Mar 2018, 12:57.
Last edited by dave13 on 24 Jul 2018, 11:04, edited 3 times in total.
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Re: Arithmetic Progression, number of terms, sum of terms etc  [#permalink]

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New post 26 Mar 2018, 12:47
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dave13 wrote:
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 :)


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
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Re: Arithmetic Progression, number of terms, sum of terms etc  [#permalink]

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New post 01 Apr 2018, 07:44
Gladiator59 wrote:
dave13 wrote:
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 :)


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


Hey Gladiator59 thanks for your comments :)

What is the difference between first consecutive integers and consecutive integers ? :?
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Re: Arithmetic Progression, number of terms, sum of terms etc  [#permalink]

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New post 01 Apr 2018, 08:09
1
dave13 wrote:

Hey Gladiator59 thanks for your comments :)

What is the difference between first consecutive integers and consecutive integers ? :?


Hey dave13,

From our prev discussion you were asking the difference between the two formulas for first n integers... the difference between those two is simply where is the starting point of our AP?

dave13 wrote:
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 :)


First n positive integers would imply d =1 and a = 1 when thinking in terms of an AP and the formula for sum of first n terms in such a case is given by \(\frac{n(n+1)}{2}\)

However, in the next case of n consecutive integers the starting point could be anything ( 1 or not)

The formula for sum of n terms of this is given by a = a & d = 1 -> in the sum formula given above. \(\frac{n}{2} (2a+(n-1)d)\)

Also for a = 1 & d = 1 ... both these formulas are one and the same as shown in prev post (in quotes below)
Gladiator59 wrote:

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.


Hope that makes sense!

Best,
Gladi
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Re: Arithmetic Progression, number of terms, sum of terms etc  [#permalink]

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New post 27 Apr 2018, 17:40
1
dave13 wrote:
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 :)
.
.
5. HOW TO FIND NUMBER OF TERMS FROM A TO B

\(\frac{first..term +last..term}{2} +1\)
.
.
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


to be continued

I will add more useful formulas later :)

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:
5. HOW TO FIND NUMBER OF TERMS FROM A TO B

\(\frac{first..term +last..term}{2} +1\)

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!
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Re: Arithmetic Progression, number of terms, sum of terms etc &nbs [#permalink] 27 Apr 2018, 17:40
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