How to Find the Nth Term of an Arithmetic Sequence

How to Find the Nth Term of an Arithmetic Sequence Nth term of an arithmetic sequence, so here in our problem we need to find the given term in each arithmetic sequence. So for A we have to find the 20th term in this sequence, for B the 11th term and for C the 8th term. Let's take a look at A, so here we have to find the 20th term. So to find the given term we need to look for the pattern in the sequence. So a subscript, so here we set up a chart and we have our term number A1 which is a subscript. And subscripts are used to show the positions of terms in the sequence. So the first term A1 is written as A1 and A2 would be written as A2 and so on. So A1 is a subscript of one which means our first terms which is five in our sequence so A2 is taking about the second term in our sequence which is eight. And then we go down. So now we have to look and find the common difference. So the common difference in this term or in this sequence is three. So we’re going to put D is our common difference. So D is our common difference. So in this sequence our common difference is D because if I take the difference between eight and five I get three, 11 – 8 is 3 and 14 – 11 is 3 and so on. So our common difference is three. So D is three here. For your first term you have no pattern. Now, for the second term 1(3) is added right because when we go to move to our second term we’re adding three. So we start off. So were adding three to our A1 so we have A5 plus 1 because it’s one in the sequence. We move down one term and we've added three so when you add three to A1 you get eight and that’s how our term is eight. For the third term, two 3’s are added right. So our third term is 1(2) so we have two 3’s that are added because when you move down to our third term we’ve added two 3’s, so that’s why it’s 2 x 3. And the number of three’s added is one less than the term number which means we’ve added. So this is actually our third term, A3 is our third term both added two 3’s so it actually would be its N – 1. So N is the term minus one. So the pattern shows that for each term the number of three’s added is one less than the term number. Okay or N – 1. So the Nth term An of an arithmetic sequence with common difference D and first term A1 is. So this is the formula. So we’re taking the Nth term of An so our An is equal to our first term A1 plus just like we we’re doing here plus N our term, the Nth term minus one times D our common difference. So to find the 20th term we need to substitute. So N is our term. So we’re looking for the 20th term. So N would be 20. A1, is our first term which would be five. And D is the common difference which we said was three. So we’re going to go ahead and just plug this in. So we have A20 equals A1, which is 5 plus N which is 20 minus one times three. So then we get A20 = 5 + 19 x 3 + 57, A20 = 5 + 57 = 62. So our 20th term is 62. Let's move on to our second set here so 11th term in this sequence. Again let’s use our formula that we just created in our first problem. So the Nth term An in the arithmetic sequence is with the common difference D and the first term A1. So An, the Nth term would be A1 + N – 1 and D is the common difference. So D is the common difference of this consecutive term. So when we find the common difference we’re taking the difference between negative four and negative two which is two. And then zero and negative two would be two and two and zero would be two, so our common difference which is D is two. So now we’re going to substitute, so the 11th term, this is giving us our N right, our Nth term. So N is 11, A1 is our first term which should be negative four and D is our common difference. So we’ll go ahead and plug it in, so A11 equals A1 is -4 plus N which is 11 minus one times two. So go ahead and solve this so A11 = -4 and 11 -1 is 10 times two is 20, A11 = 16 so the 11th term is 16. Let's go ahead and look at our last problem here. So here we have the 8th term, A1 = -4, D the common difference is two and we need to use the graph. So in this problem we need to first find the number values of several terms of the sequence and then represent the sequence on a graph. So our first term A1 is -4, so I want to have A and here’s our term right. So A1 = -4 soon we know that our second term A2 would be -4 + 2 = -2 and then we go in so on. So A3 isn’t negative two, our previous, -2 + 2 = 0. And our fourth A4 is zero, what our previous term was 0 + 2 = 2. So now we have our plots on our graph. So we have N and A. So this would be our points. So when N is one, A is -4 so that’s our first point. So this would be two and negative two and so on. So we went ahead and graph it. So you see here is one and negative one, two and negative two, here’s three and zero, four and two. So the graph points that are aligned for the sequence and visually the value of the next term can be predicted. So we can go ahead and predict it, so if we look for the 8th term and as eight we go to eighth on the X axis where it crosses. So here eight crosses at 10 on the Y axis. So our sequence of the N would be eight and A would be 10. So we see that when a number N is eight the value eight sub N right is 10. So A8 is 10 which means that the 8th term is 10. So remember that a sequence is said to be arithmetic if the difference between one term and the next term is always the same. So remember the sequences had to be arithmetic if the difference between one term and the next term is always the same. Subscripts are used to show the positions of the terms and the sequence. The first term is A1, the second term which is written as A1 and the second term is A2, A3 and so on. The Nth term An of an arithmetic sequence with the common difference D and the first term is A1. So it’s An = A1 + N – 1 x D.