Question

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Tell whether each situation produces an arithmetic or geometric sequence. Give the common difference or ratio.
1. A house gains $4,500 in value each year.
2. A clock Loses 30 seconds each hour.
3. The number of bacteria in a pond triples each day.
4. A sequence has a common difference of -1/2 and a first term of 125. Find the next three terms.

Answer 1

ok so an arithmetic sequence when there is a common difference. A Geometric sequence is when there is a common ratio. So in number 1. each year the house gains $4,500\[4,500(1+1)\]\[4,500(1+2)^{2}\]\[4,500(1+3)^{3}\]
and so on. The last number in the equation represents the number of years. The difference would be 4,5000

Answer 2

oh and 1. was an arithmetic sequence. Ok for this lets use a table.
from here we can see that the difference from one term to another is not the same, but the ratio is. so to find the ratio put the seconds of the seconds in an hour then simplfy \[\frac{ 30 }{ 3600 }=\frac{ 1 }{ 120 }\]
Making this a geometric sequence with the ratio 1/120

Answer 3

3. This does not say that it triples from the original amount but from the amount of bacteria produced in that day. So we can already tell that the difference is not going to be the same, but the ratio is making it a geometric sequence. The equation to find the ratio being:\[a=a, c=ar, b= ar ^{2}\]\[\frac{ c }{ b }=\frac{ ar }{ a }=r\]\[\frac{ b }{ c }=\frac{ c }{ a }=c ^{2}\]\[c=\sqrt{ab}\]\[c _{1}=ar, c _{2}=ar ^{2}, c _{3}=ar ^{3}... c _{k}=ar ^{k} b _{k}=ar ^{k+1}\]
C being the days with ar^k+1 being the total bacteria produced after bk amount of days. The ratio from this would be \[\frac{ ar ^{k+1} }{ c }\]

Answer 4

4. The question says the difference is -1/2 which tells you right away that this is going to be an arithmetic sequence. The challenge is that we're given a ratio to find an equal difference between the terms. This can be solved by changing the ratio to a decimal -0.5 The definition of an arithmetic sequence is\[a _{k+1}=a _{k}+d\] The number \[d=a _{k+1}-a _{k}\] is the common difference of the sequence. The formula of the nth term of an arithmetic sequence is \[a _{n}=a _{1}+(n-1)d\]or\[a _{n}=a _{k}(n-k)d\] where n and k are positive integer and d is the common difference. Since the common difference is given I think the first formula would be easier to use. \[a _{1}=125\] Since 125 is the first term and we used a sub 1 to represent it we will us a sub 2 to find the second term, a sub 3 for the third term and a sub 4 for the fourth term. Well start with trying to find the second term. \[a_{2}= 125+(2-1)-.5\]\[a_{2}=125+(-1-(-.5))\]\[a _{2}=125+(-.5)\]\[a _{2}=124.5\]\[a _{3}=125+(3-1)-.5\]\[a _{3}=125+(-1.5-(-.5))\]\[a _{3}=125-1\]\[a _{3}=124\]\[a _{4}=125+(4-1)-.5\]\[a _{4}=125+(-2-(-.5))\]\[a _{4}=125+(-1.5)\]\[a _{4}=123.5\]
So the next three term in the sequence would be:
124.5, 124, 123.5, ...