Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle. The sum of the first n terms of a sequence is n squared, plus n, plus one. That is a quadratic in n, just like the sum of an A P usually is. So this sequence is an A P, right? Hmm, hold that thought. Today is Part C of our problem bank: concept checks, short statements to judge true or false. Each one hides a small trap, and statements like these come up often in JEE, okay? So, let's begin.

0:34The three tests we will use

So, before the statements, three quick tests from this unit. First, an A P, an arithmetic progression: a, b, c are in A P if and only if two b equals a plus c. The double arrow, pointing both ways, means if and only if. Okay? Now a G P, a geometric progression. What is the test there? Yes, exactly! b squared equals a times c; the small two up top means squared. And remember, the terms of a G P must be non-zero. Right? And an H P, a harmonic progression: a, b, c are in H P when one over a, one over b and one over c are in A P, okay? Every statement today uses these three tests. Clear?

1:24Statement 1: a constant sequence

Statement one. Take a constant sequence: k, k, k, and so on, with k not zero. The three dots mean and so on. True or false: it is an A P, a G P and an H P, all three at once? Let's test. Is it an A P? Each term minus the previous one is k minus k, zero. So d, the common difference, is zero. So yes! Got it? Now, is it a G P? Each term divided by the previous one is k over k, which is one. So r, the common ratio, is one. Yes again! Okay? And an H P? The reciprocals, one over k, one over k, and so on, are again constant: an A P with d equal to zero. So the statement is true: all three! Right? But tell me, why does the statement say non-zero? Correct! The terms of a G P must be non-zero. And an H P needs reciprocals, but one over zero does not exist. Small word, big difference. With me?

2:32Statement 2: the H.M. and the G.M.

Statement two. H M is the harmonic mean, and G M is the geometric mean. True or false: the H M of two unequal positive numbers is less than their G M? True! Let's see why. Capital A, capital G and capital H stand for the A M, the G M and the H M of the two numbers. For two positive numbers, capital G squared equals capital A times capital H, okay? Now, the numbers are unequal, so capital A is strictly greater than capital G; that sign means greater than. Look at G times G equals capital A times capital H; the dot means times. Capital A is bigger than G, so capital H must be smaller than capital G; that is the sign for less than. Clear? And if the two numbers were equal? Then all three means are equal; the means are equal exactly when the numbers are equal. So the word unequal really matters. And positive matters too: we compare these three means only for positive numbers. Got it?

3:39Statement 3: an infinite G.P. with r=-1

Statement three says: an infinite G P, with common ratio r equal to minus one, has a sum. First, let's see what many students do. Okay? They remember S infinity, capital S with a small infinity sign below: a over one minus r. Put r equals minus one: a over one minus minus one, that is a over two. So the sum is a over two, right? No, no, careful! Write it out: a minus a, plus a minus a, and so on. The partial sums go a, zero, a, zero, and so on, and never settle on one number. So there is no sum. The statement is false. With me? See, a over one minus r holds only when mod r is less than one; these vertical bars mean mod. Here mod r is exactly one, so the formula does not apply. Always check mod r first. Right?

4:34Example 1: if a, b, c are in AP, are 2^a, 2^b, 2^c in GP

Now Example one, also from Part C of this chapter. True or false: if a, b, c are in A P, then two to the power a, two to the power b and two to the power c are in G P? The small letters up top are the powers. Pause the video and try it. Done? Okay, let's check. a, b, c in A P means two b equals a plus c. Clear? Now, what must we check, for the three powers to be in G P? Yes, exactly! The middle one squared must equal the product of the outer two. So two to the power b, whole squared, should equal two to the power a, times two to the power c. Okay? Two to the power b, squared, is two to the power two b. But two b is a plus c! So we get two to the power a plus c, which is two to the power a, times two to the power c. True. Perfect!

5:32Example 2: is Sn = n^2 + n + 1 the sum of an AP

Now Example two, a little harder, and it is our puzzle from the start. S n, capital S with a small n below, is the sum of the first n terms. Here S n equals n squared, plus n, plus one. Is the sequence an A P? Pause the video and try it. Done? Okay, let's check. T n, capital T with a small n below, is the n-th term: S n, minus S n minus one, where the small n minus one below is just the term number. Work it out: two n. Okay? So T n equals two n. Then T one is two, T two is four, T three is six: an A P with d equal to two, right? No, no, careful! This is the trap. T n equals S n, minus S n minus one, only for n greater than or equal to two; that is what this sign means. The first term is S one itself: one plus one plus one, three. So T one is three, not two! Got it? So the sequence is three, four, six, eight, ten, and so on. Four minus three is one, but six minus four is two. The differences are not equal, so the answer is no. It is not an A P! Right? But look, from T two onwards, four, six, eight, it is an A P with d equal to two. The A P starts only from the second term. Clear? And here is the quick test, and the answer to our puzzle. Remember, a sequence is an A P exactly when S n is capital A n squared plus capital B n, with constants capital A and capital B: no constant term. Our S n has a plus one at the end, and that spoils the first term. Puzzle solved! With me?

7:24Quick revision: four traps in AP, GP, HP and Sn

So, quick revision. One: a constant non-zero sequence is an A P, a G P and an H P, all three. Two: for two unequal positive numbers, G squared equals capital A times capital H, and the H M is less than the G M, which is less than the A M, the arithmetic mean. Three: S infinity exists only when mod r is less than one. Four: T one is S one, and T n equals S n minus S n minus one only from n equals two. Now try the three practice problems on this page. Read every word of a statement, and no trap will catch you. All the best!