Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two x, plus eight over x squared, where x is positive. Put x equals one: two plus eight, that is ten. Put x equals two: four plus two, that is six. So, can it ever go below six? Hmm. You could try a hundred values and still not be sure, right? By the end of this class, you will prove the answer in a few lines, with no calculus at all. This is Set one of our mixed problems on sequences and series, and questions like these come up often in JEE Main. So, let's begin.

0:38Set 1: what is inside

So, here is what Set one holds. Eight solved problems, all in the JEE Main pattern. Short and sharp, okay? They cover the A P and the G P, the means, special sums, and even a recurring decimal, written as a fraction. All mixed up, just like the real exam. Right? Today we take two tools from the earlier chapters, and then two problems from this set: an easy one first, then a harder one. With me?

1:06Tool 1: the A.P. formulas

Tool number one, from Chapter one: the arithmetic progression, or A P for short. Each term minus the one before it is a constant, d, the common difference. And small a is the first term. Okay? Now tell me. The n th term, counted from the first: is it a plus n times d? No, no, careful! The first term is a itself, with no d added. The second term has one d, the third has two. So the n th term has n minus one copies of d. We write it T n: capital T, with a small n at the bottom right. T n equals a, plus n minus one, times d. Clear? Next, the sum of the first n terms, S n: capital S with a small n. S n equals n over two, times, in square brackets, two a, plus n minus one times d. Got it? And if you know the last term, small l, then S n is simply n over two, times a plus l. Same sum, two forms, so one can check the other. Right?

2:07Tool 2: A.M. ≥ G.M. for a least value

Tool number two, from Chapter three. Take n positive numbers, a one up to a n: the letter a with a small number below it, and the dots mean and so on. On the left, their arithmetic mean, the A M: the sum over n. On the right, the geometric mean, the G M: the n th root of the product, with the small n sitting on the root sign. And this sign, greater than with a line below it, is read greater than or equal to. So, the A M is always at least the G M, for positive numbers. Okay? Now, when does the A M become exactly equal to the G M? Yes, exactly! When all the numbers are equal. Then, and only then. Clear? So here is the rule for a least value. For the least value of a sum, make the product of the terms constant. Then the G M is fixed, and the sum can never go below n times it. With me? And always two checks. All the terms must be positive, and equality must really be possible, that is, the terms can all become equal. Miss either one, and the answer goes wrong. Got it?

3:17Example 1: S20 of an AP with 5th term 16, 12th term 37

Now, the first problem from the set, an easy one. The fifth term of an A P is sixteen, and the twelfth term is thirty seven. Find S twenty, the sum of the first twenty terms. Pause the video, and try it first. Done? Okay, let's check. First, the fifth term. That is a plus five d, right? Oh no, careful! Tool one says n minus one. So the fifth term is a plus four d, and the twelfth term is a plus eleven d. T five is sixteen, and T twelve is thirty seven. Okay? Two equations, two unknowns. What is the fastest move here? Yes, subtract them! The a cancels. Eleven d minus four d is seven d, and thirty seven minus sixteen is twenty one. So seven d equals twenty one, and the double arrow means, so, d equals three. Right? Back to the first equation. a is sixteen minus four times three; the dot on screen means times. So a equals four. Clear? Now S twenty. Twenty over two is ten. Inside the square brackets: two times four is eight, and nineteen times three is fifty seven. So, ten times sixty five: six hundred and fifty. Got it? Always check, with the other form. The last term, T twenty, is four plus nineteen times three: sixty one. Ten times, four plus sixty one, is ten times sixty five: six hundred and fifty again. Perfect!

4:51Example 2: least value of 2x + 8/x^2 for positive x

Now, remember our puzzle from the start? Let's finish it. Find the least value of two x, plus eight over x squared, for x greater than zero. Pause the video, and try it. Hint: use tool two. Done? Okay, let's check. It is a sum of two positive terms, and we want the least value. So, just use the product of these two terms directly, right? No, no, wait! Look at that product. Two x times eight over x squared is sixteen over x. There is still an x in it, so it is not constant. The bound would move with x and tell us nothing. That is the trap. With me? So, how can we make the x cancel? The second term has x squared below, but the first term, two x, has only one x. Yes, exactly! Split two x into x plus x. Now we have three terms: x, x, and eight over x squared. Same sum. Okay? Now the product: x times x times eight over x squared. The x squared cancels, and we are left with eight. Constant! Clear? Three positive terms, so we can use the A M and G M inequality. Their sum over three is at least the cube root of their product, so the sum is at least three times the cube root of eight; the small three on the root sign means cube root. The cube root of eight is two, so the sum is at least six. Right? But is six really reached, or is it just a floor we never touch? Equality needs all three terms equal: x equals eight over x squared. So x cubed, x to the power three, is eight, and x equals two. Got it? So the least value is six, reached at x equals two. Puzzle solved! In the hook, x equals two gave exactly six, and now we have proved that nothing goes below it. Perfect!

6:50Quick revision: AP formulas and splitting terms for AM-GM

So, quick revision. One: in an A P, T n is a plus, n minus one times d, and when two terms are given, subtract them to get d. Two: S n is n over two, times two a plus n minus one times d, or n over two times a plus l, and one form checks the other. Three: for a least sum, use the A M and G M, and split the terms so the product is constant. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!