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