### Video Transcript

An arithmetic progression starting at π and ending at π has a common difference equal to π. What is the second term from the end?

In this question, weβre given some information about an arithmetic progression. Weβre told that the starting value of this arithmetic progression is equal to π and the ending value of this arithmetic progression is equal to π. Weβre also told that the common difference is equal to π. We need to use this information to determine the second term from the end of our arithmetic progression. To do this, letβs start by recalling exactly what we mean by an arithmetic progression. We recall an arithmetic progression or arithmetic sequence is a sequence where the difference between any two consecutive terms will always be the same.

So, in an arithmetic progression, the important thing to notice is the difference between any two consecutive terms in the sequence has to be the same. And the value of this difference has a name; itβs called the common difference of our arithmetic progression. For example, if we have an arithmetic sequence which starts at a value of four and to get the next term in our sequence we add a value of three and we can keep adding this value of three to generate more terms β letβs generate five terms in this sequence β then, because the difference between any two consecutive terms in our sequence is equal to positive three, we can call this value of positive three the common difference of our arithmetic sequence. And really, what this is actually telling us is to get from one term in our arithmetic sequence to the next, all we need to do is add a value of three.

But there is another way of looking at this. What if we started at our value of 16 and we wanted to find the term before this? Well, we knew to get from the fourth term to the fifth term, we needed to add the value of the common difference, we needed to add a value of three. So this means to get from 16 to 13, weβre going to need to do the opposite. Weβre going to need to subtract the value of three. And this is a useful result. It means we can subtract the common difference from our terms to go backwards in our sequence. Letβs now try and apply this to the arithmetic progression given to us in the question.

First, weβre told that our arithmetic progression is going to start at a value of π. And weβre also told that our arithmetic progression ends at a value of π. And finally, weβre also told the common difference of our arithmetic sequence is equal to π. And remember, the common difference of an arithmetic sequence will be the difference between any two consecutive terms in our sequence. For example, to find the second term in this sequence, weβre going to need to add the value of the common difference to our first term. Weβre going to need to add a value of π. So the second term in our arithmetic progression is going to be π plus π.

We can then do the same to find the third term in our arithmetic sequence. We just need to add a value of π to the second term in our sequence. If we add the value of π to π plus π, we get π plus π plus π, which is equal to π plus two π. However, we have a problem. We canβt just keep adding values of π to our terms to find the value of π because we donβt know how many times we need to add the value of π to get to our value of π. So instead, weβre going to want to use our trick of starting at our last term and finding the term before this.

Since we know to get from the second to last term to our last term, we need to add the value of the common difference of π. Then, to reverse this process, weβre going to need to subtract the value of π from our last term π. In other words, the second to last term in this sequence mustβve been π minus π because when we add the π to π minus π, we get the last term in our sequence π. And this is our final answer. Therefore, we were able to show for an arithmetic progression starting at a value of π, ending at a value of π, and has a common difference of π, then we can express the second to last term of this arithmetic progression as π minus π.