Scribe Post

May 8th, 2007

Today, we were back in the trailor and the ACs were fixed, so we weren't "frying". We started the class off with a POD, like we normally do.

POD
Q: A 50 foot ramp with one end resting on the ground, climbs 30 feet. The ramp is supported every 2.5 feet by a verticle post. What is the combined length of all the posts?

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A:
Step 1: Find the legnth of x.
30^2+x^2=50^2
900+x^2=2500
(x^2)^(1/2)=1600^(1/2)
x=40 ft.

Step 2: Find the number of posts.
length of side x/lenght between each post=number of posts
40/2.5=16

Step 3: Find T1
height of ramp/number of posts=T1
30/16=1.875

Step 4: Solve for S16 (or sum of first 16 posts)
Sn=n/2(T1+Tn)
S16=16/2(1.875+30)
S16=8(31.875)
S16=255 ft (final answer)

OR (Jaime suggested:)

Follow steps 1 & 2 and then:

Step 3:
Assume that there are 17 posts and the first one is 0ft. Therefore, you only have to do 3 steps.
Sn=n/2(T1+Tn)
S17=17/2(0+30)
S17=8.5(30)
S17=255 ft (final answer)

Then, we all decided that we would like to practice a problem that would be similar to the ones on the up-coming quiz.

Q: S is an arithmetic series such that: T6=10, T20=124.8 Find S15

A:
Step 1: Find the number of jumps.
20-6=14 jumps

Step 2: Find the difference between each term.
T20-T6=total added rate/number of jumps=d
124.8-10=114.8/14=8.2(this is the difference)

Step 3: Find T1.
6-1=5 jumps from T1 to T6
5*8.2=41
T6-41=T1
10-41=-31
T1=-31

Step 4: Find T15.
Tn=-31+(n-1)8.2
T15=-31+(15-1)8.2
T15=-31+(14)8.2
T15=83.8

Step 5: Find S15.
Sn=n/2(T1+Tn)
S15=15/2(-31+83.8)
S15=15/2(52.8)
S15=396 (final answer)

After we finished discussing this problem, we started talking about the "difference analysis" technique to find an explicit formula. We were given the sequence: 4,9,18,31,48,69,94
Then, Mr. A showed us a sequence we were already firmiliar with: 1,3,6,10,15,21
While he was showing us this new technique, the bell rang and we were not able to finish it, so I will let Naki finish it tomorrow, since she is tomorrow's scribe.