# Arithmetic sequence problem

1. May 7, 2005

### jai6638

Hey... I'd appreciate it if someone could verify these answers...

Q1) An object is dropped from an airplane. During the first second, the object falls 4.9 m. During the 2nd second, it falls 14.7 m. During the third second, it falls 24.5 m. During the fourth second, it falls 34.3 meters. If this pattern continues, how far will the object fall during the tenth second? Find the total distance the object will fall after 10 seconds.

A) An=A1+(n-1)D
A10=4.9+(10-1)9.8
A10=93.1 m
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Q2) In early growth of an embryo, a human cell divides into two cells, each of which divides into two cells and so on. The number an of new cells formed after the nth division is An= 2^(n-1). Find the sum of the first 9 terms of the series to find the total number of news cells after the 8th division.

A) 9*(2+9/2) = 9* ( 5.5 ) = 49.5 cells
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Q3) The legnth o fthe first loop of a spring is 20 inches. The length o fthe second loop is 9/10 of the length of the first loop. The length of the third loop is 9/10 of the length of the second loop, and so on. if the spring could have infinitely many loops, would it length be finite? if so, find the length.

A) how do i do this?

Last edited: May 7, 2005
2. May 7, 2005

### p53ud0 dr34m5

isnt question three a geometric series?
$$\sum_{n=0}^{\infty}20(\frac{9}{10})^n$$
if thats the case, what do you know about geometric series that makes them converge or diverge?

3. May 7, 2005

### HallsofIvy

Staff Emeritus
Actually, they are all geometric series, not arithmetic series.

4. May 7, 2005

### The Bob

This is correct. It is an arithmetic series as:
14.7m - 4.9m = 9.8m
24.5m - 14.7m = 9.8m
34.3m - 24.5m = 9.8m

You could have found that the ratio between the first and second was 3 or $$\frac{3}{1}$$. The ratio between the third and second is $$\frac{5}{3}$$. As you can see (and by doing more of them) the ratios can be illustrated by $$\frac{2k + 1}{2k - 1}$$ however this is not a common ratio so cannot be used in the geometric series $$u_k = a r^{k-1}$$ when u is the number of the kth term, a is the first term and r is the common ration (which we do not have).

$$u_k = a + (k-1)d$$ as d is the common difference (which is 9.8)

$$u_{10} = 4.9 + 9.8(10-1)$$

$$u_{10} = 4.9 + (9.8 \times 9)$$

$$u_{10} = 4.9 + 88.2$$

$$u_{10} = 93.1 \ metres$$

This is incorrect. There should be 256 cells. The sum of the first 9 terms is the same as 28 as it must divide 8 times, which is the power. A way to look at it is that one cell divides once to form two cells (so 2 is a2). Once it has divided eigth times there will be 256 cells.

I think the answer is going to be $$20 \times (\frac{9}{10})^{\infty}$$ or somthing like that. I can tell you that by adding all of the little pieces together it is getting close to 200 inches but I do not think this would help.

EDIT: I apologise to p53ud0 dr34m5. I did not understand your post when I first saw it (as I scanned it ). However what you have given is the sum to infinity (which is 200 inches). This should tell jai6638 what is the nature of the series but not what the length of the finite is. I believe we are both saying there is a finite length but it is no calculable due to the nature of the series.

Hope some of this helps.

Last edited: May 7, 2005
5. May 7, 2005

### jai6638

A3) S= 20/(1-9/10)
S=20/(1/10)
S=200 inches.
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A2) An=2^(n-1)
An=2^(9-1)
An=2^8
An=256

damn.. q2 was easy..Although, i dont undersatnd how the length in q3 can be finite if the number of loops are infinite... is it cause the length remains the same in each loop so it doesnt matter how many times the loop occurs?

Thanks much for your help guys.. appreciate it :)

6. May 8, 2005

### The Bob

This is the sum of infinity. This is what the all the lengths of the loops added together will try and add to.

Well take any of the lengths of the loops any distance down. It will be infinite. This means that one of them in infinity will be finite. However, the length can only be given in a formula form (as in my post above).