Arithmetic Means and Common Terms

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. One list goes seventeen, twenty one, twenty five, and so on, up by four. Another goes sixteen, twenty one, twenty six, and so on, up by five. Both pass through twenty one. And then they meet again at forty one, sixty one, eighty one, always twenty apart! Why twenty, and not four or five? Hold that thought. Today: arithmetic means, and the common terms of two A Ps. Both come up often in JEE problems. Let's begin!

0:35The arithmetic mean of two numbers

So, first, what is a mean? Put one number, capital A, between a and b, so that a, capital A, b is an A P, okay? Then the common difference is the same on both sides: capital A minus a equals b minus capital A. Now you tell me. What is capital A, in terms of a and b? Yes, exactly! The double arrow means it follows that. Two capital A equals a plus b, so capital A equals a plus b, over two. This is the arithmetic mean, the A M, of a and b: just their average. Simple, right?

1:13Inserting n arithmetic means between a and b

Now, why stop at one? Put n numbers between a and b so that the whole list is an A P, okay? These are the n arithmetic means, written as capital A with a small number below: A one, A two, up to A n. The three dots mean and so on. With a and b at the ends, that is n plus two terms. With me? So, here is my question. Which term number is b, and how many steps of the common difference d take you there from a? Term number n plus two, correct! To get there from a, you add d, n plus one times. So b equals a plus, n plus one, times d. Okay? Now solve for d: d equals b minus a, over n plus one. And the k th mean, capital A with a small k below, is a plus k times d: it sits k steps after a. Clear? Let's try it. Insert four arithmetic means between three and twenty three. The gap is twenty, and we want four means, so d is twenty over four, which is five, right? No, no, careful! This is the classic trap. Four means make five gaps, not four: n plus one, remember? So d equals twenty three minus three, over four plus one: twenty over five, which is four. Got it? So the four means, shown in bold, are seven, eleven, fifteen and nineteen. Each one is four more than the one before. That's it!

2:40The sum of the n means

Now, the sum of all the means. The first mean, A one, is a plus d, one step after a. The last mean, A n, is b minus d, one step before b. Okay? So what is A one plus A n? It is a plus b, yes! The plus d and the minus d cancel. Same for A two plus A n minus one, with n minus one written small, and so on. Each pair, front with back, adds up to a plus b. Right? Pair them all up, front with back: each pair gives a plus b, and n means make n over two pairs. If n is odd, the middle mean is left alone, but it is exactly a plus b over two, half a pair, so the count still works. So A one plus A two, up to A n, equals n times, a plus b over two. The small dot in the middle means times. With me? Now look: a plus b, over two, is the single A M of a and b. So the sum of the n means is n times, in brackets, that single A M, okay? Remember this; it saves a lot of work. Let's check with our four means. Seven plus eleven plus fifteen plus nineteen is fifty two. And four times, three plus twenty three over two, is four times thirteen: fifty two again. Perfect!

3:58Common terms of two A.P.s

Now the second idea. Take two A Ps with common differences d one and d two: d with a small one, and d with a small two, written below. Okay? Suppose some number is in both lists. To stay in the first, you jump by multiples of d one; to stay in the second, by multiples of d two. So what is the smallest jump that works for both? The least common multiple, exactly! We write it l c m of d one and d two, in round brackets. So the common terms form an A P too, with common difference the l c m. Right? And where does this new A P start? From the first common term. So: find the first common term, then keep adding the l c m. That's the whole method. Got it?

4:43Example 1: n means between 7 and 71 with sum 1287

Alright, example one. Some number n of arithmetic means are inserted between seven and seventy one, and their sum is one thousand two hundred and eighty seven. Find the value of n, okay? Pause the video and try it. Done? Okay, let's check. Many students start by finding the common difference d. But do we even need d here? No, we don't! This is the shortcut toppers use. The sum of the n means is n times the single A M, okay? Seven plus seventy one is seventy eight, and seventy eight over two is thirty nine. So the sum is thirty nine n. With me? So thirty nine n equals one thousand two hundred and eighty seven. Divide by thirty nine, and n equals thirty three. Just one line. See how fast that was?

5:38Example 2: common terms of 17, 21, 25, ... and 16, 21, 26

Now, let's finish our puzzle. How many terms are common to seventeen, twenty one, twenty five, and so on up to four hundred and seventeen, and sixteen, twenty one, twenty six, and so on up to four hundred and sixty six? And what is their sum? Pause the video and try it. Done? Okay, let's check. The first list goes up by four, so d one is four; the second by five, so d two is five. And twenty one is in both lists, so it is the first common term. Okay? Now, the common terms form an A P with difference l c m of four and five, which is twenty. Twenty one, forty one, sixty one, and so on. That is why the two lists meet every twenty. Mystery solved! Got it? Now, where do the common terms stop? At the bigger last term, four hundred and sixty six, right? Oh no, careful! Every common term must be in both lists, and the first list ends at four hundred and seventeen. So we stop there, at the smaller last term. Clear? So the common terms are twenty one plus twenty k, with k from zero. This must be less than or equal to four hundred and seventeen. So twenty k is at most three hundred and ninety six, and k is at most nineteen point eight. But k is a whole number, so k is at most nineteen. With me? So k runs from zero to nineteen. How many values is that? Not nineteen, but twenty: zero counts too! So, twenty common terms. Right? Now the sum. Twenty terms of twenty one give four hundred and twenty. Adding zero to nineteen gives nineteen times twenty over two, so the twenty k parts give twenty times that: three thousand eight hundred. Total: four thousand two hundred and twenty. Done!

7:27Quick revision: d = (b - a)/(n + 1), sum of means, lcm rule

So, quick revision. For n arithmetic means between a and b, d equals b minus a, over n plus one, not over n. Their sum is n times, a plus b over two. And common terms of two A Ps go up by the l c m of d one and d two, from the first common term to the smaller end. Now go and try the practice problems on this page. All the best!