The population of a city was
five-sevenths of a million in 2010 and five million in 2016. The population growth can be
described as an arithmetic sequence. Find its common difference, which
represents the annual growth of the population.
In this question, we’re told that
we have an arithmetic sequence. This is a special type of sequence
that has a common difference between each term. For example, the sequence two,
seven, 12, 17, and so on has a common difference of five, as we add five each time
to get the next term. The general term of any arithmetic
sequence, written 𝑎 sub 𝑛, is equal to 𝑎 sub one plus 𝑛 minus one multiplied by
𝑑, where 𝑎 sub one is the first term in the sequence and 𝑑 is the common
difference. It is this common difference that
we are trying to calculate in this question.
In this question, we are told that
the initial population in 2010 was five-sevenths of a million. So, we will let 𝑎 sub one, the
first term, be five-sevenths. We are also told that the
population in 2016 was five million. We can see that this would
correspond to the seventh term in our sequence. This means that 𝑎 sub seven is
equal to five. We can now substitute our values
into the formula for the general term. The seventh term five is equal to
five-sevenths plus seven minus one multiplied by 𝑑. The right-hand side simplifies to
five-sevenths plus six 𝑑.
Noting that we can write five as 35
over seven, when we subtract five-sevenths from both sides of our equation, we get
thirty sevenths is equal to six 𝑑. We can then divide both sides of
this equation by six, giving us 𝑑 is equal to 30 over 42. Both the numerator and denominator
of our fraction are divisible by six. As 30 divided by six is five and 42
divided by six is seven, our fraction simplifies to five-sevenths.
We can therefore conclude that the
common difference which represents the annual growth of the population is
five-sevenths of a million.