A guide to Number Sequences (Math)

rafae

I've seen many people asking about them. I'll post this for you guys. If you have any problem, feel free to post here, or private message me.

Linear Difference
For a linear difference, the formula goes as such,
T=An + B

A=Difference.
'n'=the number of which you are solving
Tn= the number of 'n'

I'll explain it look:

n=..1, 2, 3, 5, 6
Tn=2, 3, 4, 5, 6

So here it would be, take any number from 'n'
If i take 'n' as 1, it would be
Tn= An + B
A=difference so since the difference between 2 or 3 is 1, 4 or 5 is 1, 5 or 6 is 1, then we will take 1.
If i take 'n' as 1, then the Tn of 1 is (2) so Tn = 2
2= (1)(1) + B
B=1
So you have the value of B now which is 1.
Now put it into the equation.
Tn= n +1

Okay, here's the one for quadratic equation.
This is a bit more complex.
An² + Bn + C= Tn

The principal is the same except for a few changes.

A= difference, but this time, the difference will be divided by two. It will ALWAYS be divided by two.
B & C are both unknowns, and therefore you will obtain two equations. You will have to solve them simultaneously. If you have any problems with solving the equation, then just private message me.
number: [1,] [ 2,] [3,] [4,] [5]
Tn=.......[4,] [7,] [12,] [19,] [28]

Okay this gets a bit complex. You have to find out the second difference.
So:
4, 7, 12, 19, 28
(3...5...7...9)
(....2...2...2)

What I have written in brackets are the differences between each 'Tn'
So you have a difference of 2 in the second difference.
In quadratic equations, you always divide the difference which is 'A' by 2.
So A=2/2, =1

First equation:
An² + Bn + C= Tn
I'm taking the first number now.
1(1)² + B(1) + C=4
1+B+C=4

(i)B+C=3 (Equation one)

An² + Bn + C= Tn
1(2)² + B(2) + C = 7
4 + 2B + C = 7
(ii)2B + C= 3 (Equation two)

Now, simultaneously

B=3-C (from equation one)

2(3-C) + C= 3
6- 2C + C = 3
6 -C = 3
C=3
B=0
Just solve it simultaneously.
Now put it into the equation
n²+3, there you go.

This Quadratic Difference is not included in our syllabus..

hahahahahahaha

nice

rafae

It is in our syllabus, adnan -.-

VelaneDeBeaute

I'd lyk to add that 4 any linear distance v can use the formula..!
1st term X (n-1) X difference b/w the terms.
thats pretty easy..!

rafae

Ghalya001 said:
I'd lyk to add that 4 any linear distance v can use the formula..!
1st term X (n-1) X difference b/w the terms.
thats pretty easy..!

Yeah, that can work too

zain786

the quadratic one isnt in our syllabus (4024) i checkd

zain786 said:
the quadratic one isnt in our syllabus (4024) i checkd
yeah..

dragoninja98

Dude u guys are like such bright students....i wonder i am the only lamo sitting here....we never did linear differences !

I've seen many people asking about them. I'll post this for you guys. If you have any problem, feel free to post here, or private message me.

Linear Difference
For a linear difference, the formula goes as such,
T=An + B

A=Difference.
'n'=the angel numbers of which you are solving
Tn= the number of 'n'

I'll explain it look:

n=..1, 2, 3, 5, 6
Tn=2, 3, 4, 5, 6

So here it would be, take any number from 'n'
If i take 'n' as 1, it would be
Tn= An + B
A=difference so since the difference between 2 or 3 is 1, 4 or 5 is 1, 5 or 6 is 1, then we will take 1.
If i take 'n' as 1, then the Tn of 1 is (2) so Tn = 2
2= (1)(1) + B
B=1
So you have the value of B now which is 1.
Now put it into the equation.
Tn= n +1

Okay, here's the one for quadratic equation.
This is a bit more complex.
An² + Bn + C= Tn

The principal is the same except for a few changes.

A= difference, but this time, the difference will be divided by two. It will ALWAYS be divided by two.
B & C are both unknowns, and therefore you will obtain two angel number 733 equations. You will have to solve them simultaneously. If you have any problems with solving the equation, then just private message me.
number: [1,] [ 2,] [3,] [4,] [5]
Tn=.......[4,] [7,] [12,] [19,] [28]

Okay this gets a bit complex. You have to find out the second difference.
So:
4, 7, 12, 19, 28
(3...5...7...9)
(....2...2...2)

What I have written in brackets are the differences between each 'Tn'
So you have a difference of 2 in the second difference.
In quadratic equations, you always divide the difference which is 'A' by 2.
So A=2/2, =1

First equation:
An² + Bn + C= Tn
I'm taking the first number now.
1(1)² + B(1) + C=4
1+B+C=4

(i)B+C=3 (Equation one)

An² + Bn + C= Tn
1(2)² + B(2) + C = 7
4 + 2B + C = 7
(ii)2B + C= 3 (Equation two)

Now, simultaneously

B=3-C (from equation one)

2(3-C) + C= 3
6- 2C + C = 3
6 -C = 3
C=3
B=0
Just solve it simultaneously.
Now put it into the equation
n²+3, there you go.
13-year-old post and still so relevant!!! Thanks a lot, buddy