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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 24 Mar 2018, 12:57
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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

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

where \(n\) is number of terms

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



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

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



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


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 :)
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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
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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
Re: Arithmetic Progression, number of terms, sum of terms etc   [#permalink] 01 Apr 2018, 08:09
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