Given that 𝑥, 𝑎, 𝑏, 𝑦 is an arithmetic sequence, evaluate 𝑦 minus 𝑥 over 𝑏 minus 𝑎.
Because we know these four variables form an arithmetic sequence, we can say that 𝑎 is halfway between 𝑥 and 𝑏. We can also say that 𝑏 is halfway between 𝑎 and 𝑦. We can say this because in an arithmetic sequence, each consecutive term is one common difference away from the previous term. But another way to say this would be that 𝑎 is the arithmetic mean of 𝑥 and 𝑏.
If we know that the arithmetic mean is the sum of values divided by the number of values and we’ve already said that 𝑎 was the arithmetic mean of 𝑥 and 𝑏, then 𝑎 equals 𝑥 plus 𝑏 divided by two. And since 𝑏 is the arithmetic mean between 𝑎 and 𝑦, 𝑏 is equal to 𝑎 plus 𝑦 divided by two.
To solve 𝑥 minus 𝑦 over 𝑏 minus 𝑎 would be to rewrite these equations so that 𝑥 is by itself and 𝑦 is by itself. We’ll move our 𝑏-equation over to give ourselves a little bit more room. For equation one, we multiply both sides of the equation by two, which gives us two 𝑎 equals 𝑥 plus 𝑏. And when we subtract 𝑏 from both sides, we see that two 𝑎 minus 𝑏 equals 𝑥.
For our second equation, we’ll solve for 𝑦. By multiplying both sides of the equation by two, we get two 𝑏 equals 𝑎 plus 𝑦. And then, we subtract 𝑎 from both sides, which gives us two 𝑏 minus 𝑎 equals 𝑦. This means we’re ready to plug in what we found for 𝑥 and for 𝑦 into our equation. 𝑦 is two 𝑏 minus 𝑎, and our numerator is 𝑦 minus 𝑥. 𝑥 is two 𝑎 minus 𝑏, and the denominator would stay the same, 𝑏 minus 𝑎.
To solve, our first step will be to distribute the subtraction, which gives us two 𝑏 minus 𝑎 minus two 𝑎 plus 𝑏 in the numerator. We can combine our like terms. We have two 𝑏 plus 𝑏 equals three 𝑏. And negative 𝑎 minus two 𝑎 equals negative three 𝑎, which will give us three 𝑏 minus three 𝑎 over 𝑏 minus 𝑎. In our numerator, both the 𝑏 and the 𝑎 have a factor of three. We can undistribute that three so that the numerator becomes three times 𝑏 minus 𝑎, which is all over 𝑏 minus 𝑎. 𝑏 minus 𝑎 divided by 𝑏 minus 𝑎 equals one. So, 𝑦 minus 𝑥 over 𝑏 minus 𝑎 equals three.
Before we finish, let’s consider one other method for solving this question. When we’re working with an arithmetic sequence, we know that there is a common difference between each consecutive terms. And this means we could rewrite 𝑎 in terms of 𝑥, its previous term, and the common difference 𝑑. We could say that 𝑎 equals 𝑥 plus 𝑑. And when we think about our third term, it’s equal to our second term 𝑎 plus the common difference.
However, we can also write 𝑏 in terms of 𝑥 and 𝑑. We can say that 𝑏 will be equal to the first term 𝑥 plus two 𝑑. So, we have 𝑏 equals 𝑥 plus two 𝑑. And of course, this means that we could write the fourth term as 𝑥 plus three 𝑑. We could say that 𝑦 equals 𝑥 plus three 𝑑. And then, we take what we found for 𝑦 and plug it in to our equation. Our numerator becomes 𝑥 plus three 𝑑 minus 𝑥, and our denominator is 𝑏 minus 𝑎. We plug in 𝑥 plus two 𝑑 for 𝑏 and 𝑥 plus 𝑑 for 𝑎.
In our denominator, we need to distribute this subtraction. Minus 𝑥 plus 𝑑 is the same thing as minus 𝑥 minus 𝑑. And we see in our numerator that 𝑥 minus 𝑥 equals zero. And in our denominator, 𝑥 minus 𝑥 equals zero. In our denominator, we have two 𝑑 minus 𝑑 which will just leave us with one 𝑑; we can write that as 𝑑. And in the numerator, we have three 𝑑. Three 𝑑 over 𝑑 simplifies to three and confirms that 𝑦 minus 𝑥 over 𝑏 minus 𝑎 equals three.