Arithmetic progression

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VeeN22

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A computer tablet infected with a malware virus has resulted in its flat rectangular screen being affected by 1/3 of the screen being blocked on the first day. On each successive day it blocks a further portion of the screen, as big as a 1/3 of the area it blocked on the previous day.
The malware programme continues to act indefinitely. Before this infection, the area of the flat screen is 1000 square units.

If the malware programme continues to act indefinitely, what fraction of the user's screen will be eventually blocked out?
 
This is not an arithmetic progression. What kind is it?

Use that fact to write an expression for the progression, and show us your work. Then we can work with you to find the answer.
 
Dr.Peterson said:
This is not an arithmetic progression. What kind is it?

Use that fact to write an expression for the progression, and show us your work. Then we can work with you to find the answer.
Click to expand...

Okay, it's a geometric progression, I worked it out as follows:
Common ratio is 1/3
To get my terms I calculated 1/3×1000 and continued multiplying with 1/3 until I got:
1000/3; 1000/9; 1000/27; 1000/81;

1000/243 and on.

Now I don't know at which fraction the screen will eventually be blocked out completely.
 
On day \(n\) we have visible the following portion of the screen:

[MATH]V_n=\left(\frac{2}{3}\right)^n[/MATH]
And so the amount blocked is:

[MATH]B_n=1-V_n=1-\left(\frac{2}{3}\right)^n[/MATH]
 
VeeN22 said:
Okay, it's a geometric progression, I worked it out as follows:
Common ratio is 1/3
To get my terms I calculated 1/3×1000 and continued multiplying with 1/3 until I got:
1000/3; 1000/9; 1000/27; 1000/81;

1000/243 and on.

Now I don't know at which fraction the screen will eventually be blocked out completely.
Click to expand...
You haven't defined "it". What does your sequence represent? Read carefully.

The problem said,

On each successive day it blocks a further portion of the screen, as big as a 1/3 of the area it blocked on the previous day.​

So your sequence represents, not the total amount blocked on a given day, but the additional amount blocked each day, beyond what was blocked the day before. So the total amount blocked is the sum of your series. You need to find whether (and if so, when) this sum exceeds 1.

I think MarkFL has misread the problem, as saying that each day 1/3 of what was unblocked will be blocked (so that 2/3 of what was unblocked remains unblocked), rather than 1/3 of the previous day's advance.
 
Yes, I misread...to the corner with me...LOL!
 
Dr.Peterson said:
You haven't defined "it". What does your sequence represent? Read carefully.

The problem said,

On each successive day it blocks a further portion of the screen, as big as a 1/3 of the area it blocked on the previous day.​

So your sequence represents, not the total amount blocked on a given day, but the additional amount blocked each day, beyond what was blocked the day before. So the total amount blocked is the sum of your series. You need to find whether (and if so, when) this sum exceeds 1.

I think MarkFL has misread the problem, as saying that each day 1/3 of what was unblocked will be blocked (so that 2/3 of what was unblocked remains unblocked), rather than 1/3 of the previous day's advance.
Click to expand...

So if you're saying it blocks a further portion on each successive day, does it mean that I should add 1/3 to the first term?
 
The portion being blocked on day \(n\) is:

[MATH]B_n=\left(\frac{1}{3}\right)^n[/MATH]
And so the total being blocked on day \(n\) is the sum:

[MATH]T_n=\sum_{k=1}^{n}\left(\frac{1}{3}\right)^k[/MATH]
Hence:

[MATH]3T_n=\sum_{k=1}^{n}\left(\frac{1}{3}\right)^{k-1}=\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}=1+T_n-\frac{1}{3^n}[/MATH]
What do you get upon solving for \(T_n\)?
 
MarkFL said:
The portion being blocked on day \(n\) is:

[MATH]B_n=\left(\frac{1}{3}\right)^n[/MATH]
And so the total being blocked on day \(n\) is the sum:

[MATH]T_n=\sum_{k=1}^{n}\left(\frac{1}{3}\right)^k[/MATH]
Hence:

[MATH]3T_n=\sum_{k=1}^{n}\left(\frac{1}{3}\right)^{k-1}=\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}=1+T_n-\frac{1}{3^n}[/MATH]
What do you get upon solving for \(T_n\)?
Click to expand...

I got 1/3 for Tn
 
Just to be more clear, the equation is:

[MATH]3T_n=1+T_n-\frac{1}{3^n}[/MATH]
 
You are close...it is:

[MATH]T_n=\frac{1}{2}-\frac{1}{2\cdot3^n}[/MATH]
What does this imply as \(n\) grows without bound?
 
That's what "grows without bound" means: the sequence continues forever.

The question is, what happens to the value of [MATH]T_n[/MATH] when n gets very large? For example, what will the hundredth term be? How about the millionth? Will the whole screen ever be covered?
 
Dr.Peterson said:
That's what "grows without bound" means: the sequence continues forever.

The question is, what happens to the value of [MATH]T_n[/MATH] when n gets very large? For example, what will the hundredth term be? How about the millionth? Will the whole screen ever be covered?
Click to expand...

Tn increases an n increases but only when n is up to 100. Any number beyond that is undefined. The last fraction I got is 1/2. Maybe that's the answer
 
Why do you say that it is undefined? Did you just try calculating it with a calculator? That's not the best thing to do here.

If so, try thinking instead. When n gets large, what happens to [MATH]3^n[/MATH]? So what happens to the whole expression?
 
Dr.Peterson said:
Why do you say that it is undefined? Did you just try calculating it with a calculator? That's not the best thing to do here.

If so, try thinking instead. When n gets large, what happens to [MATH]3^n[/MATH]? So what happens to the whole expression?
Click to expand...

As n gets larger and larger, so does 3n and the whole expression too.
 
Well, if you divide 1 by a very large number, what will the result be? Is it very large?
 
In fact, the result of the division becomes very small, right? In fact, you can make it as close to zero as you want.

So in [MATH]T_n=\frac{1}{2}-\frac{1}{2\cdot3^n}[/MATH], the second fraction approaches zero (and is always positive), so [MATH]T_n[/MATH] approaches [MATH]\frac{1}{2}[/MATH], but never reaches or goes above that. So, as you guessed, that is the answer: eventually, just under half of the screen is blocked.

What we've done here is called a limit; have you learned anything about that concept yet?
 
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