Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
6. Geometric Sequences
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Exercise 51 Page 338

Practice makes perfect
a We first need to convert between gallons and fluid ounces. To do so we will use the conversion factor.
128 fl oz/1 galMultiplying 16 gallons by this conversion factor will convert it to fluid ounces.
16 gal * 128 fl oz/1 gal
â–Ľ
Simplify
16 gal *128 fl oz/1 gal
16 *128 fl oz/1
2048 fl oz/1
2048 fl oz
Now we can write a sequence that represents amount of the soup in fluid ounces left after each day. Since 14 of the soup is served, 34 of the soup is left after each day. Therefore, after the first day 1536 fluid ounces of soup is left. 2048 * 3/4= 1536 Let's multiply this number by 34 each time to find the first five terms.

The first five terms of the sequence are 1536, 1152, 864, 648, and 486.

b We see that there is a common ratio of 34 between each pair of consecutive terms. Since a_1= 1536 and r= 34, we can write an equation that represents the nth term of this geometric sequence.

a_n= a_1( r)^(n-1) ⇔ a_n= 1536( 3/4)^(n-1)

c We know that graphs of geometric sequences form an exponential curve when the common ratio is positive. If this common ratio is between 0 and 1, the graph looks like the graph of an exponential decay function.

Recall that the function is never equal to zero since the x-axis is the horizontal asymptote. Similarly, geometric sequences with a common ratio between 0 and 1 cannot be zero. As a result, we can theoretically claim that the soup will never run out. However, in real life, eventually there will not be enough soup to serve.