Sums of a G.P.

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. You drop a ball from ten metres. Every time it lands, it bounces back to three fifths of the height it fell from, again and again, forever. Infinitely many bounces! So the total distance must be infinite, right? Hmm, are you sure? Today we learn the sums of a geometric progression, a G P, and you will find that distance yourself. These sums come up often in JEE questions. So, let's begin.

0:33The sum of the first n terms

So, first the notation. Here, a is the first term and r the common ratio. S n, a capital S with a small n below it, is the sum of the first n terms, counted from the first term. So S n equals a, plus a times r, plus a times r squared, and so on, up to a times r to the power n minus one. The small raised numbers are powers. Okay? Now a lovely idea. Multiply the whole sum by r. What happens to each term? Yes, exactly! Each term becomes the next one. So r times S n equals a times r, plus a times r squared, and so on, up to a times r to the power n. Almost the same terms! Right? So, subtract the first row from the second. Everything in the middle cancels; only a times r to the power n, and a, survive. So r times S n, minus S n, equals a times r to the power n, minus a. With me? Take out S n and a, then divide by r minus one. S n equals a times, r to the power n minus one, over r minus one. The equals sign with a slash through it means not equal to: r is not equal to one. Clear? But why do we need r not equal to one? What goes wrong when r equals one? Correct! The bottom becomes zero, and we cannot divide by zero. But when r is one, every term is just a. So S n is a, added n times: n times a. That's it!

2:09The sum to infinity

Now, the big question. Can we add infinitely many terms? Take r equals one half: r squared is one quarter, r cubed one eighth, and r to the power ten, one over one thousand and twenty four. So what happens to r to the power n as n keeps growing? Yes, exactly! It gets closer and closer to zero. This happens whenever mod r is less than one: the two vertical bars mean the modulus, the size of r without its sign. The arrow means tends to, and the sideways eight is infinity. Okay? Now put that into our formula for S n. r to the power n becomes zero, so the top becomes a times, zero minus one: minus a. And minus a over r minus one is a over one minus r. Right? So S infinity, a capital S with the infinity sign below it, equals a over one minus r. But only when mod r is less than one. The condition is part of the formula, okay? Let me test you. Take r equals two, with a positive first term. The formula gives a over one minus two: minus a. So the sum is minus a, right? No, no, careful! The terms keep doubling, all positive. How can they add to a negative number? When mod r is greater than or equal to one, the terms do not shrink, so there is no finite sum. Always check mod r first. Got it?

3:37Topper's trick: recurring decimals

Now, a trick that toppers use. The bar over thirty seven means three seven repeats forever. So zero point three seven recurring is zero point three seven, plus zero point zero zero three seven, and so on. Each term is one hundredth of the one before: an infinite G P! Okay? So tell me. What is the first term, and what is the common ratio? Correct! The first term is zero point three seven, and r is zero point zero one. Mod r is less than one, so S infinity gives zero point three seven over one minus zero point zero one, that is, over zero point nine nine. Multiply top and bottom by one hundred: thirty seven over ninety nine. Nice, right? When only some digits repeat, like zero point one two three with the bar over two three only, take one hundred and twenty three, minus the part that does not repeat, one. Below, two nines for the two repeating digits, then one zero for the one that does not: nine hundred and ninety. So, one hundred and twenty two over nine hundred and ninety, or sixty one over four hundred and ninety five. Clear? The same idea handles seven, plus seventy seven, plus seven hundred and seventy seven, and so on. First, the number made of k nines is ten to the power k, minus one. Right? So the number made of k sevens is seven ninths of that: seven over nine, times, ten to the power k minus one. And those powers of ten, ten, a hundred, a thousand, form a G P, so our S n formula finishes the job. With me?

5:19Example 1: sum of 1 + 3 + 9 + ... to 8 terms

Alright, example one. Find the sum of one, plus three, plus nine, and so on, to eight terms. Pause the video and try it. Done? Okay, let's check. The first term a is one, r is three over one, that is, three, and n is eight. Okay? Now, r is three, bigger than one. Is that a problem for our formula? No problem at all! The condition mod r less than one is only for S infinity. So S eight, S with a small eight below, equals one times, three to the power eight minus one, over three minus one. Right? Three to the power eight is six thousand five hundred and sixty one. Minus one: six thousand five hundred and sixty. Over two: three thousand two hundred and eighty. That's it!

6:13Example 2: total distance of a bouncing ball

Now, our bouncing ball. Let's finish it. The ball is dropped from ten metres; the small m on screen means metres. Each time, it rebounds to three fifths of the height it fell from. Find the total distance travelled. Pause the video and try it. Done? Okay, let's check. After the ten metre drop, the first rebound is three fifths of ten, six metres; the next, three point six metres, and so on. So the heights form a G P with r equal to three fifths, zero point six. Okay? So the total is just S infinity: ten over one minus zero point six, twenty five metres, right? Oh no, careful! The ball falls the first ten metres only once. But every rebound height is travelled twice: up, then down. With me? So the total is ten, plus two times, in brackets, six plus three point six, and so on. That bracket is an infinite G P with first term six and mod r less than one, so S infinity works. Clear? Six over one minus zero point six is six over zero point four: fifteen. Two times fifteen is thirty, plus ten: forty metres. Infinitely many bounces, only forty metres! Puzzle solved. Infinitely many terms can have a finite sum, when mod r is less than one. Got it?

7:40Quick revision: Sn, sum to infinity, recurring decimals

So, quick revision. S n equals a times, r to the power n minus one, over r minus one, for r not equal to one, and n times a for r equal to one. S infinity equals a over one minus r, only when mod r is less than one. And recurring decimals, and sums like seven plus seventy seven, become G P sums. You did really well today. Now go and try the practice problems on this page. All the best!