Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Start with four. Take away two, you have two. Add one, three. Take away a half, two and a half. And keep going, forever, each new term half the last one, with the sign flipped. Infinitely many terms. Can the total ever settle on one exact number? Strange, right? By the end of this class, you will find that number in one line. This is Part A of our sequences and series problem bank, at JEE Main level, and sums to infinity like this come up often. So, let's begin.

0:41Part A: a 30-minute test

So, here is what Part A holds. Eight solved problems, all at JEE Main level. Short, sharp questions, okay? And they are mixed: A P, G P, means, telescoping and special sums, all together. Nobody tells you which chapter a question comes from, just like the real exam. Right? The plan: finish Chapters one to five, then attempt Part A as one thirty minute test. So today, two tools that some of these problems need, then two of its problems, one easy and one harder. With me?

1:18Tool 1: the sum of a G.P. to infinity

Tool number one, from Chapter two: the sum of a G P to infinity. Small a is the first term, and r is the common ratio, each term divided by the one before it. So r equals T two over T one: the second term over the first. The little two and one written low are term numbers, and the line between T two and T one is a fraction bar. Okay? Now the formula. S infinity, that is S with a small infinity sign written low, equals a over one minus r. That's it, just one line. Right? But wait. Does this formula work for every G P? Try r equals two: the terms double every time. Can they ever add up to a finite number? No, never! They keep growing, without end. So the formula works only when mod r is less than one. The two straight bars around r mean modulus: the size of r, ignoring its sign. The pointed sign means less than, so r lies between minus one and one. Then the terms shrink, and the sum settles down. Got it? And if mod r is greater than or equal to one, that is the sign with a line under it, then S infinity does not exist at all. Many students forget this check. Always do it first. With me?

2:35Tool 2: A.M. ≥ G.M.

Tool number two, from Chapter three: the A M, G M inequality. For two positive numbers a and b, the arithmetic mean, a plus b over two, is greater than or equal to the geometric mean, the square root of a times b. The tick-shaped sign is the square root. Okay? Multiply both sides by two: a plus b is at least two root a b. And equality holds exactly when a equals b. Remember that part; we will need it. Right? Now think. We want the least value of a sum, a plus b. For two root a b to be one fixed number, what must stay constant? The product, yes, exactly! For the minimum of a sum, make the product of the terms constant. Then two root a b is just a number, and that number is your bound. Clear? And two checks, every single time. The numbers must be positive, and equality must be possible. If equality can never happen, the bound is not the least value.

3:35Example 1: sum to infinity of 4 - 2 + 1 - 1/2 + ...

Got it? Now, remember our puzzle from the start? Let's finish it. Find four, minus two, plus one, minus a half, and so on, to infinity. The three dots mean the terms go on forever. Pause the video, and try it first. Hint: use tool one. Done? Okay, let's check. The first term, a, is four. The common ratio is the second term over the first: minus two over four, that is, minus a half. See, r is negative; that is why the signs keep flipping. With me? Now, before using any formula, what must we check first? The condition, yes! Mod r is one half, and one half is less than one. So S infinity exists, and the jumping total really does settle down. Perfect! So, S infinity is four over one minus a half, which is eight, right? No, no, careful! This is the trap. Here r is minus a half, so one minus r is one minus, in brackets, minus a half. Minus times minus gives plus: one plus a half. Always keep the sign of r. Okay? One plus a half is three over two. Four over three halves is four times two over three; the cross means times. That is eight over three. So infinitely many terms, and the total settles at exactly eight over three. Puzzle solved!

5:04Example 2: least value of x + 1/x + y + 1/y

Now a slightly harder one. Find the least value of x plus one over x, plus y plus one over y, where x and y are both greater than zero; the pointed sign there means greater than. Pause the video, and try it. Hint: use tool two. Done? Okay, let's check. We have four positive terms, and tool two says: make the product constant. So which terms should we pair together? x with one over x, yes, exactly! Their product is just one. So, by tool two, x plus one over x is at least two root of x times one over x; the small dot means times. The root of one is one, so x plus one over x is at least two. Right? In exactly the same way, y plus one over y is at least two. Clear? Add the two: the whole sum is at least two plus two, which is four. Okay? So, is the least value four? Wait. Remember the second check. Can the sum actually equal four? Yes, it can! Equality needs x equal to one over x, and y equal to one over y. The double arrow means implies: with x and y positive, x equals y equals one. One plus one plus one plus one is four. The bound is attained. Correct! And why did the question say that x and y are positive? What if x could be negative, say minus one? Then x plus one over x is minus one, plus one over minus one: minus two. Less than two! The bound breaks completely. That is why the A M, G M inequality needs positive numbers. Got it? So, the least value is four, at x equals y equals one. Pair the terms, write the bound, check equality: three moves, and the problem is done. See how fast that was?

7:03Quick revision: infinite GP condition and AM GM equality

So, quick revision. One: S infinity equals a over one minus r, only when mod r is less than one, and keep the sign of r. Two: the A M, G M inequality needs positive numbers, and for the least value of a sum, make the product constant. Three: always check that equality is possible. Now go and attempt all eight problems on this page as one thirty minute test, then try the practice problems. You will do well, I promise. All the best!