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Wanahton purified a portion of water with 900 grams of contaminants. Each hour, a third of the contaminants was filtered out.

Let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? Then, write an explicit formula for the sequence, g(n)=?

Answer :

abidemiokin

Answer:

g is a GEOMETRIC SEQUENCE

Explicit formula g(n) = [tex]900(\frac{1}{3})^{n-1}[/tex]

Step-by-step explanation:

If wanahton purified a portion of water with 900 grams of contaminants and each hour, a third of the contaminants was filtered out then the amount of grams filtered out first hour will be 1/3 of 900grams = 300grams.

The amount filtered out in the second hour will be 1/3 of 300grams = 100grams.

Given the initial amount of contaminant = 900g

Contaminant after 1st hour = 300g

contaminant after 2nd hour = 100g and so on

It is seen that the amount of contaminant keep reducing by one third each hour. The sequence formed will be 900, 300, 100... which is a GEOMETRIC SEQUENCE because they are reducing by the same factor each term which is 1/3.

Given the geometric sequence 900, 300, 100...the nth term of the sequence is expressed as shown;

g(n) = [tex]ar^{n-1}[/tex]

r is the common ratio = [tex]\frac{300}{900}= \frac{100}{300}[/tex] =  [tex]\frac{1}{3}[/tex]

n is the umber of terms

a is the first term which serves as the initial value of the contaminant= 900

Substituting the values given to get the explicit formula for the sequence we have;

g(n) = [tex]900(\frac{1}{3})^{n-1}[/tex]

The amount of contaminants (in grams) that remained by the beginning of the nth hour is [tex]900(\frac{1}{3})^{n-1}[/tex]

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