Find 𝑎 sub 45, given the arithmetic sequence from 18, 26, 34, all the way through to 698, where 𝑛 is greater than or equal to one.
We’ll begin by recalling what we mean by an arithmetic sequence. It’s a number sequence in which the difference between one time and the next is a constant. In this question, we’re looking to find 𝑎 sub 45. In other words, we’re looking to find the value of the 45th term in the sequence. Now, one method we have is to write the sequence out, adding the common difference each time until we reach the 45th term. There are two reasons this might be problematic though. Firstly, it’s a very long-winded method. And secondly, with that number of calculations, we’re increasing the probability that we make a mistake.
So instead, we’ll begin by calculating the 𝑛th term for our sequence. This is the rule that will allow us to find any term in our sequence given the term number. And there are two ways that we can calculate this. We’ll look at both of them. The first involves using a formula. Now, this might be written in a number of ways. But we’re going to write it as 𝑎 sub 𝑛 equals 𝑎 sub one plus 𝑛 minus one times 𝑑. Now, it makes a lot of sense that 𝑎 sub one would be the first term because 𝑛 is the term number. 𝑑, however, is the common difference between each term.
The first three terms of our sequence are 18, 26, and 34. We can see that the first term in our sequence is 18. So we say 𝑎 sub one is 18. The common difference is found by subtracting each term from the term in front. 26 minus 18 is eight. And 34 minus 26 is eight and so on. So the common difference 𝑑 is eight. Substituting what we know into our formula, we can say that 𝑎 sub 𝑛 is equal to 18 plus 𝑛 minus one times eight. Convention generally dictates that we put the eight in front of the 𝑛 minus one. And so we have an 𝑛th term. It’s 𝑎 𝑛 equals 18 plus eight times 𝑛 minus one.
The question is asking us to find 𝑎 sub 45. That’s the 45th term. So we let 𝑛 be equal to 45. And we find 𝑎 sub 45 is 18 plus eight times 45 minus one, which is 18 plus eight times 44. Now, eight times four is 32. So eight times 40 is 320. 320 plus 32 is 352. So we find 𝑎 sub 45 is equal to 18 plus 352, which is equal to 370. And so, by finding the 𝑛th term using the formula 𝑎 sub 𝑛 equals 𝑎 one plus 𝑛 minus one 𝑑, we found the 45th term. 𝑎 sub 45 is equal to 370.
Now, the second method takes a slightly different approach. We still begin by finding the difference between the terms. And once again, we find that’s eight. This looks a little bit like the eight times tables. So the first part of our 𝑛th term is eight 𝑛. We then write out the eight times tables above the terms in our sequence. So the first term is eight. The second is 16. And the third is 24.
We then notice that to get from each term in the eight times tables to our sequence, we need to add 10. So our 𝑛th term, 𝑎 sub 𝑛, is equal to eight 𝑛 plus 10. Once again, we substitute 𝑛 equals 45 into our formula. So we get eight times 45 plus 10. And eight times 45 is 360. So 𝑎 sub 45 is 360 plus 10, which is 370. Notice we’ve got the exact same answer by using a slightly different method.
𝑎 sub 45 in either case is 370.