Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. One list goes seven, twelve, seventeen, and so on: take its first five hundred terms. Another goes three, ten, seventeen, and so on: take its first three hundred terms. Both contain seventeen. So, how many numbers are in both lists? Write out eight hundred numbers and compare? That would take forever, right? By the end of this video, you will do it in a few lines. Sequences and series comes up often in JEE Main, and this chapter mixes the whole unit together. So, let's begin!

0:38How this chapter is built

So, first, where are the real exam questions? The real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to four. The short dash on screen just means to, okay? This chapter is different. Its problems are original, in the same style as those real ones. So it is all fresh practice. Right? And they come in three mixed sets: eight solved problems, then four practice problems, in each. Mixed means any part of the unit can come next, so you must spot the type yourself. Got it?

1:13What recent JEE Main questions favour: A.P.s and G.P.s

Alright, think back over the four chapters of this unit. Which types of questions do you expect recent JEE Main papers to favour? Good, hold on to your guesses, and let's see if they match. The course lists five types. Let's go one by one. Type one: sums of A Ps, where the sum of the first n terms, a partial sum, is given to you, okay? The tool? Capital T with a small n below is the n th term, and capital S with a small n below is the sum of the first n terms. T one equals S one, and T n equals S n, minus S of n minus one, for n greater than or equal to two: the sign like a greater than, with a line below. Clear? Type two: common terms of two A Ps, numbers in both lists, just like our puzzle, right? They form an A P too. Its common difference is the l c m, the least common multiple, of d one and d two, the two common differences, with a small one and two below. And it starts from the first common term. Okay? Type three: G Ps, with one condition on the product of the terms, and another on their sum. With me? Here the trick from Chapter two helps. Take three terms as a over r, a, and a times r. Multiply them: r cancels, and the product is just a cubed. So the product condition gives you the middle term, a, straight away. Nice, isn't it?

2:35What recent JEE Main questions favour: means and series

Type four: the A M, G M inequality, for extreme values, the greatest or least value of something. The short dash just joins the two names. Okay? For positive numbers a one, a two, up to a n, with small numbers below, their sum over n, the A M, is greater than or equal to the n th root of their product, the G M, right? The small n on the root sign means n th root, and the dots mean and so on. And note: positive numbers only. Using it for negative numbers is a classic trap. Now, a quick question. When does the A M actually equal the G M? When all the numbers are equal, exactly! That is the only time the bound is reached. So if the numbers can never be equal, that bound is not the answer. Got it? Type five: series summed by the method of differences, or by telescoping, where the terms cancel in a long chain. With me? If each term, T k, is f of k, minus f of k minus one, then the arrow says: the sum, the big Greek letter sigma, from k equals one to n, is just f of n minus f of zero. Everything in the middle cancels. That's it!

3:48Example 1: 10th term when Sn = 5n^2 + 3n

Now, our first example, from Set two, marked easy. If S n equals five n squared plus three n, find the tenth term. The small two up top means squared. Pause the video, and try it. Done? Okay, let's check. Many students put n equals ten straight away, and get five hundred and thirty. So the tenth term is five hundred and thirty, right? No, no, careful! S ten, capital S with a small ten below, is five times a hundred, plus thirty: five hundred and thirty. But that is the sum of the first ten terms, not the tenth term, T ten. Clear? So, use the tool from type one. T n equals S n, minus S of n minus one: five n squared plus three n, minus five times n minus one squared, minus three times n minus one. Okay? Five n squared, minus five times n minus one squared, is five times, two n minus one. And three n, minus three times n minus one, is just three. So T n equals five times, two n minus one, plus three, which is ten n minus two: an A P with common difference d equal to ten. With me? Now put n equals ten, which is at least two. Ten times ten, minus two: ninety eight! Quick check. S nine is five times eighty one, plus twenty seven: four hundred and thirty two. S ten minus S nine: five hundred and thirty minus four hundred and thirty two, ninety eight again. Perfect!

5:23Example 2: common terms of 7, 12, 17, ... and 3, 10, 17, ...

Now, remember our puzzle? Let's finish it. This one is from Set two, marked medium. How many terms are common to the first five hundred terms of seven, twelve, seventeen, and so on, and the first three hundred terms of three, ten, seventeen, and so on? Pause the video, and try it. Done? Okay, let's check. The first list goes up by five, so d one is five; the second by seven, so d two is seven. And seventeen, in both lists, is the first common term. Okay? Now, before anything else: where must the common terms stop? At the smaller last term, yes! Counting from the first term, the five hundredth term is seven plus four hundred and ninety nine times five: two thousand five hundred and two. The three hundredth is three plus two hundred and ninety nine times seven: two thousand and ninety six. Every common term must be in both lists, so we stop at two thousand and ninety six. Right? The common terms form an A P from seventeen, with difference the l c m of five and seven, thirty five. Seventeen, fifty two, eighty seven, and so on. Clear? So seventeen plus thirty five k must be less than or equal to two thousand and ninety six: the sign like a less than, with a line below. So thirty five k is at most two thousand and seventy nine, and since k is a whole number, k is at most fifty nine. Got it? So, fifty nine common terms, right? Oh no, careful! k starts from zero, which gives seventeen itself. So k runs from zero to fifty nine: sixty values. Sixty common terms! Eight hundred numbers, and we never wrote them out. Puzzle solved!

7:05Quick revision: five types, Tn from Sn, common terms

So, quick revision. Five types: partial sums, common terms, G P conditions, the A M, G M inequality, and telescoping. From S n, T n is S n minus S of n minus one, for n at least two, so S ten is not the tenth term! For common terms, step by the l c m of d one and d two, stop at the smaller last term, and count k equal to zero. Now open Set one and start solving. You are ready for this. All the best!