# Video: CBSE Class X • Pack 2 • 2017 • Question 12

CBSE Class X • Pack 2 • 2017 • Question 12

06:05

### Video Transcript

The first term of an arithmetic progression is five, the last term is 45, and the sum of all its terms is 400. Find the number of terms and the common difference of the arithmetic progression.

To begin this question, we can first write down the key information that we’ve been given. The question tells us that the first term of the arithmetic progression is five. We’ll call this term 𝑎 one, with the one representing the number of the term. The question also tells us that the last term is equal to 45. We’ll call this term 𝑎 𝑛. Again, here the 𝑛 represents the number of the term, and currently we don’t know what this is. We’re also told that the sum of all the terms is 400, and we’ll call this 𝑆 𝑛.

Now it’s really important to also note that we’re dealing with an arithmetic progression. What this tells us is that successive terms will grow or shrink by the same value each time. In other words, the difference between successive terms of our arithmetic progression will always be the same. We’ll come back to the implications of this later on.

To begin, we’re gonna focus on finding the number of terms in our arithmetic progression. In our notation, we have defined that our progression has a total of 𝑛 terms, with 𝑎 𝑛 being the last term. And we know that the sum of all of these 𝑛 terms is 400.

Now you may be familiar with the following formula, which says the sum of all 𝑛 terms in an arithmetic progression is equal to the number of terms, 𝑛, divided by two multiplied by the first term add the last term. Now for this video, we’re not gonna go into the full proof of this formula. However, you might be able to gain an intuitive understanding with a small rearrangement. Looking at this rearrangement, we can see that we have taken an average of our first term and our last term by adding them together and dividing by two.

Now earlier on, we mentioned that, for an arithmetic progression, the difference between successive terms is the same. Since the terms change by the same value each time, finding the average of the first and the last terms is essentially the same as finding the average of all the terms in the progression. We then multiply this average by the total number of terms, 𝑛. Thinking about the formula in this way, we might be able to see that multiplying the average value of each of our terms by the total number of terms will indeed give us a sum of all the terms.

Now that we understand this, let us now use this formula to find 𝑛, the number of terms. We first substitute in the values that we already know: 400 as the sum of all the terms, five as our first term, and 45 as our last term. To get rid of this factor of a half on the right-hand side of our equation, we multiply both sides by two. And we can also simplify our brackets, since five add 45 is equal to 50.

We now have that 50 times 𝑛 is equal to 800. And we therefore divide both sides of our equation by 50, to see that 800 divided by 50 is equal to 𝑛. 800 divided by 50 is 16, and we’ve therefore found that the total number of terms 𝑛 is 16.

Now that we found this, we’re able to tidy things up by replacing the 𝑛 with 16. 𝑎 16, or the 16th term of our arithmetic progression, is 45. And the sum of all 16 terms is equal to 400.

Now let’s move on to finding the common difference between our terms, which we’ll call 𝑑. Now we already know the value of our first term, 𝑎 one, is five, but what about the value of our second term, 𝑎 two? Well, since we’ve defined our common difference to be 𝑑, the value of 𝑎 two will be the value of 𝑎 one plus the common difference 𝑑.

We already know the value of 𝑎 one is five, so 𝑎 two is five plus 𝑑. By similar reasoning, the value of our third term, 𝑎 three, will be the value of the previous term, 𝑎 two, plus 𝑑 again. We can use our previous line of working to substitute in for 𝑎 two, which is 𝑎 one plus 𝑑. And we know that 𝑎 one is five, so we have that 𝑎 three is five plus two 𝑑.

Here we can begin to spot a pattern. And by following this pattern, we can build up a formula for any number term in the arithmetic progression. Here we see that, for our second term, 𝑎 two, we have the first term, 𝑎 one, plus one lot of the common difference. And for our third term, 𝑎 three, we have our first time, 𝑎 one, add two lots of the common difference. Using this pattern, we’re able to say that the term numbered 𝑘 in our arithmetic progression will have the value of our first term plus 𝑘 minus one times the common difference 𝑑.

Using this rule that we found, we should now be able to calculate the common difference, 𝑑, since we know the value of the 16th and final term of our arithmetic progression. We first substitute 𝑘 equals 16 into our formula. We know that 𝑎 16 is equal to 45 and that 𝑎 one is equal to five. 16 minus one is 15, so we now have that 45 equals five plus 15𝑑. We subtract five from both sides, to find that 40 is equal to 15𝑑, and then divide both sides by 15, to find that 40 over 15 is equal to 𝑑.

Finally, we can simplify our fraction by dividing the top and bottom half by five. And we find that the common difference, 𝑑, is equal to eight over three. We have now fully answered the question. And we found that the number of terms, 𝑛, is 16 and the common difference, 𝑑, is eight over three for the arithmetic progression.