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
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:
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 = 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 = 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
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