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

The population of a city was 8/9 of a million in 1999 and 16 million in 2016. If it can be described as an arithmetic sequence, find the population in 2019 to the nearest million given the population growth is constant.

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### Video Transcript

The population of a city was eight-ninths of a million in 1999 and 16 million in 2016. If it can be described as an arithmetic sequence, find the population in 2019 to the nearest million, given the population growth is constant.

So the key bit of information in this question is that we’re told that we can model our situation as an arithmetic sequence. And an arithmetic sequence is a sequence where there is a common difference between our terms. And when we’re dealing with an arithmetic sequence, what we have is this general form. And that is that 𝑎 sub 𝑛 is equal to 𝑎 sub one plus 𝑛 minus one 𝑑. This is where 𝑎 sub 𝑛 is any term. 𝑎 sub one is a first term. We’ve got 𝑛. That’s our term number. And then we’ve got 𝑑, which is our common or constant difference. And as well as being told that this is an arithmetic sequence, we’re also told that the population growth is constant. So that fulfils our common difference.

Okay, great. So we’ve got this. Let’s use it to try and find out what the population in 2019 is going to be. Well, first of all, we know 𝑎 sub one, our first term. And this is eight-ninths. Well, it’s actually eight-ninths of a million. But as we’re dealing with millions throughout, I’m just gonna use the eight-ninths cause it’s gonna be easier to calculate. And then we can bring back the millions when we give the final answer.

Then we also know that 𝑎 sub 17 or the 17th term is gonna be equal to 16, or 16 million. And that’s because we’re told that the population is 16 million in 2016. Okay, great. That’s the information we’ve been given. Now what’s the next step?

Well, what we can do is we can use our general form to form an equation. So we’ve got the 17th term is gonna be equal to the first term plus 17 minus one 𝑑. Okay, let’s substitute in the values we know. And when we do that, we’re gonna get 16 is equal to eight-ninths plus 16𝑑. So then what we’re gonna do is we’re gonna subtract eight-ninths from each side. And when we do that, we get 144 over nine minus eight-ninths is equal to 16𝑑. And we did that on the left-hand side, just convert it into ninths because it’s easier. So we did 16 multiplied by nine, which gives us 144.

So now we can calculate, because what we can do on the left-hand side is just subtract the numerator. And when we do that, we get 136 over nine equals 16𝑑. So then what we can do is divide both sides by 16. And when we do that, we get seventeen eighteenths, so 17 over 18, is equal to 𝑑. So this is our common or constant difference. What this means is that the population grows 17 over 18 or seventeen eighteenths of a million each year.

So have we finished the problem? Well, no, because what we want to find is the population in 2019. So therefore, if we’re looking for the population in 2019, what we’re looking for is the 20th term or 𝑎 sub 20. And this is gonna be equal to the first term, eight-ninths, plus 20 minus one, because 20 is the term number, multiplied by a common difference, which is seventeen eighteenths. So therefore, the 20th term is gonna be equal to eight-ninths plus 19 multiplied by 17 over 18. So therefore, the 20th term is gonna be equal to 18.83 et cetera.

So have we finished here? Well, no, because what we’re looking for is the answer or population to the nearest million. So therefore, we can say that the population in 2019 is going to be 19 million. And this is to the nearest million.