MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT HELP!

[hamid_ishfaq]

aoa guys I want help in understanding transformation and patterns they are just making me go crazy!!

can some one tell me an easy way to interpret graphs on which shapes are made and it says describe the transformation i just cant understand!!!

the patterns question is that how to find this nth term DAMN IT!!!

a)3,6,10,15

b)7,14,22,31

plz hellp yaar!!!!!!

i no many of you out there would be experiencing the same problems so kindly participate in the discussion to solve your problems

 

[Sijan92]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

tell me the year. sometimes sequences can be solved by seeing the entire question.

 

[hassam]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

well i was havin da same problem previously.i may not be able to explain well so go to this link.GUARANTEE U wil unrstnd each nd everthing.the sequences u gave r QUADRATIC sequence in which 2nd difference is constant.
RULE: IF U FIND THE 2ND DIFFERENCE CONSTANT n start vid n over 2 times n squared
http://www.bbc.co.uk/schools/gcsebitesi ... rev1.shtml
HERE U GO
tension finish!

 

[emkay]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

thenks man, the link was helpful

 

[hamid_ishfaq]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

tell me the year. sometimes sequences can be solved by seeing the entire question.

yaar i dnt no the year for the questions but the 1st is from the past papers while the second is self made

 

[hamid_ishfaq]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

well i was havin da same problem previously.i may not be able to explain well so go to this link.GUARANTEE U wil unrstnd each nd everthing.the sequences u gave r QUADRATIC sequence in which 2nd difference is constant.
RULE: IF U FIND THE 2ND DIFFERENCE CONSTANT n start vid n over 2 times n squared
http://www.bbc.co.uk/schools/gcsebitesi ... rev1.shtml
HERE U GO
tension finish!

thanx man the link was helpfull but still m unable ti find the pattern like for the first part it has to be 1/2 n(square) + something
now how to find this something???

 

[XPFMember]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

i replied to ur msg,check ur inbox.
Well how to find this something thats easy.U just have to make an eqn and solve it.
Lets take ur question- 3,6,10,15
now when n=1 it is 3 let that smthng be x
1/2 (n^2 + x)=that term
1/2(1^2 + x)=3
1+x=3 * 2
x=6-1= 5
so 1/2(1^2 + 5)=3
now find x if n=2
u will get it as 8
for n=3 it will be 11
so u get a new sequence for x 5,8,11 each next term is by adding 3 so nth term for x will be 3n+2 now substitute this in place of x so ur final ans. of nth term is=>1/2(n^2+3n+2).
Always check ur nth term to compare valus given u should get the same values by sustituting values for n.
hope this helps!!
JazakAllah!

 

[hamid_ishfaq]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

i replied to ur msg,check ur inbox.
Well how to find this something thats easy.U just have to make an eqn and solve it.
Lets take ur question- 3,6,10,15
now when n=1 it is 3 let that smthng be x
1/2 (n^2 + x)=that term
1/2(1^2 + x)=3
1+x=3 * 2
x=6-1= 5
so 1/2(1^2 + 5)=3
now find x if n=2
u will get it as 8
for n=3 it will be 11
so u get a new sequence for x 5,8,11 each next term is by adding 3 so nth term for x will be 3n+2 now substitute this in place of x so ur final ans. of nth term is=>1/2(n^2+3n+2).
Always check ur nth term to compare valus given u should get the same values by sustituting values for n.

hope this helps!!
JazakAllah!

man thnx lot lot infinite times it was damn easy actually i h??ad come to this x part by myslf bt wasnt able to find a nth term for this x well thnx alot but how did u gt ds +2 personal observation or by some method

 

[mehdi1028]

Re: MAN THIS TRANSFORMATION AND PATTERNS IS FREAKIN ME OUT H

for part a:
n(n - 1)/2 or n/2(n - 1)

for part b:
7n + 0.5(n-square - 3n +2 )
for e.g
5th is :
7(5) + 0.5(5-square - 3(5) +2)
=35 + 0.5(25-15+2)
=35 + 0.5(12)
=35+6
=41

I hope u guys understand!
THE FOLLOWING FORMULA CAN BE USED TO FIND Nth TERM OF ANY SEQUENCE:

a + (n-1)d + 0.5(n-1)(n-2)C

Where :
d = the first difference (7 in the sequence above).
C = The difference increase.( +7,+8,+9 so the increase is 1)
a = the first term