Alright now in the morning we had a test on probability which I'm sure most of you did well on as it seem easy, all that review and group work payed off i guess.

As for the afternoon class we started to introduce our new unit a bit more, Sequences.

A Sequence is an ordered list of numbers that follow a certain pattern or a rule

we learn a few definitions as shown on today's slides:

we work on questions such as:

8, 14, 20, 26, 32, .......

we know by looking at this that it appears to be increasing by 6

so the next 3 values would be 38, 44, 50, but what will be the 56 value? the 116?

you could count all the way up to the 56th term but that's a lot.

so we can create a formula for this sequence.

So now to find a Common Difference

to refer to the Common Difference we use d.

To find d we will need to find the value of 2 terms tn and t(n-1).

so d=tn-t(n-1)

d=t2-t1

d=14-8

d=6

now to create the formula for this Sequence.

to find the nth term in an Arithmetic Sequence. we use:

tn=a+(n-1)d

a being the first term in this case 8

n is the "rank" of the nth term in the sequence which is 56 and 116.

d as we know is the common difference.

now that we have the formula we can find t116 and t56

t116:

Solution:

a=8

d=6

n=116

t116 = 8 + (116-1)(6)

t116 = 8 + (115)(6)

t116 = 8 + 690

t116 = 698 therefore the 116th term will be 698.

t56:

Solution:

a=8

d=6

n=56 (other values have stay the same only thing different is the n term of the sequence)

t56 = 8 + (56-1)(6)

t56 = 8 + (55)(6)

t56 = 8 + 330

t56 = 338 therefore the 56th term will be 338.

to find the implicit/linear function

using: y=mx+b we can find this function

since this sequence(8,14,20,...) is constantly increase by 6 we now that this is a linear function

we can create the equation is we know:

d and the y intercept

we already know d so lets find the y-int

m=d

b=y-int

x=n

So we already know: tn=dn + b

to find b we must find the oth term to do this we know the sequence increase by 6 so from 8, the 1st term we can find the 0th term

t0=8-6

t0=2

and there we have it t0, which is the y-int is b so the equation is:

tn=6n + 2

so to test this let use n=4 which is 26

t4 = 8 + (4-1)(6)

t4 = 8 + (3)(6)

t4 = 8 + 18

t4 = 26 therefore the 4th term dose = 26 to our first equation.

now is it the same for our implicit equation?

t4=6(4) + 2

t4=24+2

t4=26 same answer with less the work.

that's pretty much what we went over this class, now its time for me to choose a scribe at random..............ok done and it will beee: wendy