I summon the Mathematicians

number two can be pretty easily solved. you start with 100%, and 10% is filtered out each time. so the amount filtered out each time is given by the recursive

Xsub1=100
Xsubn= Xsub(n-1)*90/100

i'm pretty sure the explicit formula for this is (90/100)^(n-1)= L where L is the amount of light getting through. so if we substitute 50 for L, we get (90/100)^(n-1)=50 . i'm kinda too lazy to do the math from there, but i hope that helps at least somewhat.

Here's a way of solving the first one:
A: 1+r+r^2=3/t1
B: 1+r+r^2=12/t3
C: t1=t3/r^2

Substitute C into A to get:
D: 1+r+r^2=(3r^2)/t3

The left sides of B and D are equal so the right sides must also be equal:
E: 12/t3=(3r^2)/t3

Multiply both sides by t3 to get:
12=3r^2

So r=2

Solving A for t1 yields:
t1=3/7

Nth term = t1r^(N-1)

First six terms are 3/7, 6/7, 12/7, 24/7, 48/7, 96/7

3/7 + 6/7 + 12/7 = 21/7 = 3
12/7 + 24/7 + 48/7 = 84/7 = 12

For the second one, it seems you need to know how light transmission varies with thickness.

P = Percentage of light transmitted
T = Thickness of filter in cm

P = .9^T

a) P = .9^1.3
b) .5 = .9^T

For question 1 I offer the initial sequences as
0+1+2 = 3
and
2+4+6 = 12

The only way I know to determine the Nth term is to write out the series. 0,1,2,4,6,9,12,16,20,....

I offer this solution in opposition to the previous answer because I am aware that high school problems, while frequently mind benders, are generally cooked to provide simple closed answers.

I don't have an algorithm for finding sequences. I just took a prospective teacher's test and the book I had to study from said to look at the sequence and attempt to determine a pattern. I said some impolite variation of "Gee Thanks! As if that isn't what I've always done." However, I have, over the years of playing with questions like this noticed that you can frequently figure out the sequence by a little bit of scratching on paper.

In this case I considered what numbers add to three. Well you have 1+1+1 or 0+1+2. The numbers which would add to twelve have to include the third number from the first sequence. That is 1 or 2. If it's 1 then I need two numbers to add to 11....5 and 6 are all that come to mind. My mind doesn't see anything obvious in a series of 1,1,1,5,6. The other option implies that the first number is zero, 0. 0+1+2=3; so 2 plus what = 12? or what sum gives a ten? 12-2=10. I plug in guesses. 3,7; 4,6;5,5. Series generally have to progress larger so 6,4 is a valid combination, but not a reasonable guess for a series. I wrote out 0,1,2,4,6 first. I had a solution that is relatively easy and I quit. Looking at the other posiblities, 3,7 and 5,5 when I write them out I don't see a relationship that makes obvious sense.

I should admit that I'm an engineer who has spent many years relying heavily upon the KISS principle of design and solutions. Keep It Simple Stupid. 0,1,2,4,6 is simple in my mind.

Now the next step, to the Nth term. So how are the terms in hand progressing. Nothing so obvious as adding, doubling, tripling.
Let's look at the sum between the terms. 0+1=1;1+2=3;2+4=6;4+6=10.
Hmm 1,3,6,10. Nothing immediately obvious, but let's look again.
3-1=2;6-3=3;10-6=4. I declare AHA!! there is a simple incrementing series! So the difference to the next term must be one greater than the difference from the previous term.

Sums 0+1=1;1+2=3;2+4=6;4+6=10
next compare sums 1,3,6,10
those differences are 2,3,4
thus the next sum difference must be 5
thus 1,3,6,10,15
so the next series sum = 6+5 = 11

look at it visually:

0 1 2 4 6 9 12 16 20 25 30 . . . . series you're looking for
.\/..\/.\/..\/.\/. \/. \/.. \/.. \/...\/
1 3 6 10 15 21 28 36 45 55 sum of adjacent terms
..\/.\/..\/...\/..\/..\/...\/...\/...\/
2 3 4 5 6 7 8 9 10 difference between sums of adjacent terms

This is a non-trivial brain bender. I hope you had at least fifteen minutes available to work this one alone. I made several mistakes before I worked it correctly.

Had to use dots between arrows to keep them in place when fully posted. The spaces shrank too much to maintain alignment.

Kelly

KellyS,

You appear to have overlooked the fact that the problem stated that the sequence was geometric.

Look at it this way, since every element is geometric then it increases with some power (I'm gonna assume it's a square) so

t1=x^0; t2=x^1; t3=x^3 and so in such that tn=x^(n-1)
goven the first relationship (t1+t2+t3=3) this becomes (x^0+x^1+x^2=3), putting this in a solvable form it becomes (x^2+x-2=0) and using the quadratic form to solve for x, you get x=(-1±√((-1)^2-4*1*(-2)))/2 or x=(-1±(-3))/2 or x=1 or x=-2

using the second equation as a check t3+t4+t5=12 becomes x^2+x^3+x^4=12

substitution x=1 then 1^2+1^3+1^4=1+1+1=3 and 3≠12 so x=1 is not a solution,

so try the other possibility and let x=-2 now x^2+x^3+x^4 becomes
(-2)^2+(-2)^3+(-2)^4= 4+(-8)+16=12

so the series is 1=(-2)^0, -2=(-2)^1, 4=(-2)^2, -8=(-2)^3, 16=(-2)^4, -32=(-2)^5, 64=(-2)^6 and the nth element is (-2)^(n-1)

The second problem is an application of Beer's law & E=m^t where E is transmittance and t is thickness.

If E=90% (0.9) @ t=1cm this is 0.9=m^1, therefore m=0.9, if the new filter thickness is 1.3cm the transmission E=0.9^1.3=0.871996≈88%

Thickness for 50% is a little harder since your solving for t for in E=m^t (0.5=0.9^t). Using logarithms you can treat multiplication like addition and 0.5=0.9^t becomes log(0.5)=log(0.9^t) whisb\c can be rearranged as log(0.5)=tLog(0.9) so t=log(0.5)/log(0.9)= 6.578813 cm. A check would be to put 6.578813cm into Beer's law (E=0.9^6.578813=0.5)

Rap

Rap,
You're correct also. When I got to 12=3r^2, I mistakenly assumed that r had to be positive. Both sequences are solutions.