Arithmetico-Geometric Progressions

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. One half, plus two quarters, plus three eighths, plus four sixteenths, and on forever. The numerators keep growing. So does the sum blow up to infinity, or settle down to some number? Hard to say, right? By the end of this class, you will get the exact answer in one line. Today: arithmetico geometric progressions, or A G Ps. They come up often in JEE, so let's begin!

0:30What is an A.G.P.?

So, what is an AGP? Let's build one. On the left, an A P: one, two, three, four. The three dots mean and so on. On the right, a G P: one, x, x squared, x cubed. With me? Now multiply them term by term. We get one, plus two x, plus three x squared, plus four x cubed, and so on. Now tell me: is this an A P? Or a G P? Neither, exactly! The gaps are not equal, and the ratios are not equal. Each term is an A P term times a G P term: an arithmetico geometric progression. In general, take an A P with first term a and common difference d, and multiply by one, r, r squared, and so on. Clear? So, what is the n th term? Look at the pattern. Here it is. Capital T with a small n below is the n th term. In the square brackets, the A P part: a, plus n minus one times d. Then times r to the power n minus one, from the G P, okay? Notice: n minus one in both places, not n. Got it?

1:42Summing an A.G.P.: multiply by r, subtract

Now, how do we add up an AGP? Let capital S be the sum. The trouble is the A P part: a, a plus d, a plus two d, always growing. So, how do we kill that A P part? Hint: in an A P, what stays the same from one term to the next? The common difference d, yes! So, multiply the whole of S by r. Every term moves one place to the right: now a plus d times r squared sits right below a plus two d times r squared. Right? Now subtract, matching the powers of r. First, a stays. Then a plus d, minus a, leaves d, so we get d r. Next, d r squared, and so on. After the first a, every term is just d times a power of r: a plain G P, with first term d r and common ratio r. Got it? If it runs forever, what does this G P add up to? Yes, first term over, one minus the ratio: d r over, one minus r. But only when mod r is less than one. The two straight bars around r mean modulus. And the left side is capital S times one minus r. Clear? Now divide by one minus r. S infinity, the S with the infinity sign below, equals a over, one minus r, plus d r over, one minus r, whole squared. And remember the condition: mod r less than one. Okay?

3:09Two standard results: sum of n x^n for |x| less than 1

Now, two results worth remembering. Take our first series again: one plus two x plus three x squared, and so on. Same trick, with r equal to x: subtracting x times S leaves one plus x plus x squared, and so on, which is one over, one minus x. With me? Divide by one minus x once more: one over, one minus x, whole squared. On screen, sigma, the big Greek letter, means add up. Below it, n greater than or equal to one: start from n equals one. So, sigma of n times x to the power n minus one. Okay? Multiply by x, and sigma of n times x to the power n is x over, one minus x, whole squared. Both need mod x less than one, as in the heading. Now, our puzzle! One half plus two quarters plus three eighths: that is n over two to the power n, from n equals one up to infinity. The second result, with x equal to one half. So, what is the sum? Half over, one minus half, whole squared: half over a quarter. Exactly two! And mod x is one half, less than one, so the result applies. Infinitely many growing numerators, and the sum is just two. Beautiful, isn't it?

4:26Example 1: 1 + 4/5 + 7/25 + 10/125 + ... to infinity

Now, our first example from the course. Find one, plus four fifths, plus seven over twenty five, plus ten over one hundred and twenty five, and so on, to infinity. Pause the video, and try it. Done? Okay, let's check. Numerators one, four, seven, ten: up by three each time, so a is one and d is three. Right? Denominators one, five, twenty five, one hundred and twenty five: times five each time. So r is one fifth, and mod r is less than one. Clear? First part: a over, one minus r. That is one over four fifths, which is five over four. Now the second part: d r over, one minus r, whole squared. You try it. d r is three fifths, and four fifths squared is sixteen over twenty five. So we get fifteen over sixteen. Oh no, careful: do not drop the r in d r. Using just three is a common slip. Five over four is twenty over sixteen, so S infinity is thirty five over sixteen. That's it!

5:35Example 2: sum of 1 + 2·2 + 3·2^2 + ... + 10·2^9

Now a harder one. Find one, plus two times two, plus three times two squared, and so on, up to ten times two to the power nine. The little dot means times. Pause the video, and try it. Done? Okay, let's check. Here a is one, d is one, and r is two. So, straight into the S infinity formula, right? No, no, wait! That is the trap. Here r is two, so mod r is greater than one, and there are only ten terms. The S infinity formula needs an infinite series with mod r less than one. So we use the method itself. With me? So, here is capital S, up to ten times two to the power nine. Now multiply everything by the ratio, two. Two S is the same terms, shifted one place to the right. But look at the end: an extra term, ten times two to the power ten, with no partner above it. When we subtract, what happens to it? It stays, with a minus sign! The first term, one, stays too. Everything else pairs up: two times two, minus one times two, is two; three times two squared, minus two times two squared, is two squared; and so on, up to two to the power nine. Students often drop that last term. Don't! S minus two S is just minus S, on the left. The bracket is a G P of ten terms with ratio two: two to the power ten, minus one, that is one thousand and twenty three. And ten times one thousand and twenty four is ten thousand two hundred and forty. Okay? So minus S is minus nine thousand two hundred and seventeen. The double arrow means so: capital S is nine thousand two hundred and seventeen. That's it!

7:18Quick revision: nth term, multiply by r, S infinity

So, quick revision. An AGP is an A P times a G P, term by term, with the n th term you see here. To add one up: multiply by r, subtract, sum the G P, and keep the last term! And when mod r is less than one, S infinity is a over, one minus r, plus d r over, one minus r, whole squared. Now go and try the practice problems on this page. You can do them. All the best!