Solving application problems with geometric sequences
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of
instead of
In these problems, we can alter the explicit formula slightly by using the following formula:
Solving application problems with geometric sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
Write a formula for the student population.
Estimate the student population in 2020.
The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let
be the student population and
be the number of years after 2013. Using the explicit formula for a geometric sequence we get
We can find the number of years since 2013 by subtracting.
We are looking for the population after 7 years. We can substitute 7 for
to estimate the population in 2020.
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
recursive formula for
term of a geometric sequence
explicit formula for
term of a geometric sequence
Key concepts
A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
The constant ratio between two consecutive terms is called the common ratio.
The common ratio can be found by dividing any term in the sequence by the previous term. See
[link] .
The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See
[link] and
[link] .
A recursive formula for a geometric sequence with common ratio
is given by
for
.
As with any recursive formula, the initial term of the sequence must be given. See
[link] .
An explicit formula for a geometric sequence with common ratio
is given by
See
[link] .
In application problems, we sometimes alter the explicit formula slightly to
See
[link] .
Questions & Answers
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