Thread: Geometric Sequences 2

The color of some clothing fades over time when washed. Suppose a pair of jeans fades by 5% with each washing.

a) What percent of the color remains after one washing?
95%

b) If t1 = 100, what are the first 4 terms?
t1 = 100 r = 0.95
100, 95, 90.25, 85.7375

c)What is the value of r?
r = 0.95

d) What percent of the color remains after 10 washings? I went wrong here.
tn = t1(r^n-1)
tn = (100)(0.95^10-1) I plugged 10 into n
tn = 63.0249%
The correct answer is 59.87%, I am not sure what I did wrong...

e) How many washings would it take so that only 25% of the original color remains? I also do not know how to solve this one.
tn = (100)(0.95^n-1)
25 = (100)(0.95^n-1)

I think we need to solve for n but I do not know the algebra steps to get n by itself since it is an exponent...

I would love some feedback, I felt I did d) correctly, but I guess not... Thanks for any contribution!

If t1 = 100, that means after zero washings. Some people might write "t0" for that, but that's a matter of convention. If t1 = 100, then t2 = 95 (after one washing), and t11 = 59.87, after 10 washings. You need to plug 11 in for n, not ten. Just a matter of how the counting is done.

Part e requires logarithms. If you're doing this, I suppose you're familiar with logarithms. Go back and review them. If you have problems, ask further questions.

I'm in high school grade 11, and I have actually never been taught to do logs, I'm just doing them now...

I understand d) and I've tried doing e) again:
tn = (100)(0.95^n-1)
25 = (95^n-1)
(log(25))=n-1(log(95))
1.39794001 = n-1 (1.977723601)
1.39794001/(1.977723601)= n-1
1.39794001/(1.977723601)+1 = n
n = 1.71

The answer is after 27 washes...

Your answer to 3) is correct, I think, but I don't follow your figuring. Did you maybe leave something out? For instance, I got

n = 28.027

which corresponds to 27 washings. You do need to subtract one from the value of n to get the "washing number." That's a consequence of counting the way you do. If you started counting from 0, it might be easier.

You might want to glance at the attached pdf.

You probably don't know what natural logs are yet (notation: ln). Learning about that would probably be an enrichment activity for you, not necessary for your class.

Originally Posted by Mathnood768

I'm in high school grade 11, and I have actually never been taught to do logs, I'm just doing them now...

I understand d) and I've tried doing e) again:
tn = (100)(0.95^n-1)
25 = (95^n-1)

This is incorrect. You should have 0.25= (0.95)^{n-1}

(log(25))=n-1(log(95))
1.39794001 = n-1 (1.977723601)
1.39794001/(1.977723601)= n-1
1.39794001/(1.977723601)+1 = n
n = 1.71

The answer is after 27 washes...

I presume this is the answer "in the back of the book".