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Arithmetic Progression, number of terms, sum of terms etc

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Senior Manager
Joined: 09 Mar 2016
Posts: 442
Arithmetic Progression, number of terms, sum of terms etc [#permalink]

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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
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

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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,
Senior Manager
Joined: 09 Mar 2016
Posts: 442
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

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01 Apr 2018, 07:44
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,

What is the difference between first consecutive integers and consecutive integers ?
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

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01 Apr 2018, 08:09
1
KUDOS
dave13 wrote:

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)

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