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Question Video: Solving Word Problems Involving Arithmetic Sequences Mathematics • 9th Grade

The population of a city was 5/7 of a million in 2010 and 5 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.

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Video Transcript

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.

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