The Sum of an A.P.

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Take twenty four, twenty one, eighteen, and so on, going down by three each time. Keep adding. How many terms make the sum seventy eight? Here is the surprise: there are two correct answers! Two different counts, the same sum? How? Hold that thought. Today we learn the sum of an A P, and sums like this come up often in JEE problems. So, let's begin.

0:28The sum of the first n terms

So, first the notation. S n, a capital S with a small n written below it, is the sum of the first n terms, counted from the very first term. The first term is a, and the last of these n terms is l, small l. Okay? Now write the sum forwards: a, plus a plus d in round brackets, plus a plus two d, and so on, the three dots, up to l. Right? Now the same sum backwards: l, then l minus d, then l minus two d, down to a. Put one row below the other, and add column by column. What does each column give? Yes, exactly! Each column gives a plus l. The plus d and the minus d cancel. There are n columns, so two S n equals n times, a plus l. The double arrow means it follows that S n equals n over two, times a plus l: half the number of terms, times first plus last. Got it? And if the last term is not given? Use l equals a plus, n minus one, times d. Then S n equals n over two, times, in square brackets, two a plus, n minus one, times d. Same sum, second form. Clear? Quick one. Add the first fifty odd numbers: one, three, five, and so on. What is the last one? Ninety nine, correct! So, fifty over two, times one plus ninety nine: twenty five times one hundred, two thousand five hundred. That is fifty squared! Nice, right?

1:58Spot an A.P. from its sum Sn

Now open the brackets in the second form. S n equals d over two times n squared, plus, in round brackets, a minus d over two, times n. It is a quadratic in n, with no constant term. With me? And it works both ways. The double headed arrow means if and only if: a sequence is an A P exactly when S n equals capital A n squared plus capital B n, with no constant term. And then d is two times capital A, okay? Let's test it. S n equals three n squared plus five n. So d is three, the number in front of n squared, right? No, no, careful! T n, capital T with a small n, is the n th term. It equals S n minus S n minus one, with n minus one written small, for n greater than or equal to two. That gives T n equals six n plus two. With me? And T one equals S one, which is eight, so the first term fits too. The terms are eight, fourteen, twenty, and so on. So d is six, two times three, not three. The number in front of n squared is only half of d.

3:08Topper's trick: the ratio of kth terms

Clear? Now, a trick that toppers use. Take an odd number of terms, two k minus one. The middle one is T k, with a small k. First plus last is twice the middle term, so the sum is the number of terms times the middle term: S, with two k minus one written small, equals two k minus one, times T k. Okay? Now take two arithmetic progressions whose sums of n terms are in the ratio f of n is to g of n. The colon means is to, right? Then their k th terms are in the ratio f of two k minus one is to g of two k minus one. Why? Each k th term is its sum of two k minus one terms, over two k minus one, and that factor cancels in the ratio. Clear? Try it. The sums are in the ratio seven n plus one is to four n plus twenty seven. For the eleventh terms, what do you put for n? Twenty one, yes! Two times eleven, minus one. Putting n equals eleven is the trap here. Got it? Seven times twenty one, plus one, is one hundred and forty eight. Four times twenty one, plus twenty seven, is one hundred and eleven. That cancels to four over three, so the eleventh terms are in the ratio four is to three. That's it!

4:26Example 1: sum of all three-digit numbers divisible by 7

Alright, example one. Find the sum of all three digit numbers divisible by seven. Pause the video and try it. Done? Okay, let's check. The smallest three digit multiple of seven is seven times fifteen, one hundred and five. The largest is seven times one hundred and forty two, nine hundred and ninety four. Okay? They go up by seven each time, so it is an A P: a is one hundred and five, d is seven, and l is nine hundred and ninety four. With me? Now, how many terms? Nine hundred and ninety four minus one hundred and five, over seven, is one hundred and twenty seven. So n is one hundred and twenty seven, right? Oh no, careful! That counts the gaps between the terms, and there is always one more term than gaps. So add one: n is one hundred and twenty eight. Got it? Now the sum. We know the last term, so use n over two, times a plus l. One hundred and twenty eight over two is sixty four, and one hundred and five plus nine hundred and ninety four is one thousand and ninety nine. Sixty four times that: seventy thousand, three hundred and thirty six.

5:40Example 2: terms of 24, 21, 18, ... for sum 78, two answers

Done! Now, remember our puzzle from the start? Let's finish it. How many terms of twenty four, twenty one, eighteen, and so on, must be taken for the sum to be seventy eight? Explain the double answer. Pause the video and try it. Done? Okay, let's check. Here a is twenty four, and d is minus three. No last term is given, so use the second form: n over two, times, in square brackets, forty eight plus, n minus one, times minus three, equals seventy eight. Okay? Inside the bracket, that is fifty one minus three n. Multiply by two: n times, fifty one minus three n, equals one hundred and fifty six. Rearrange and divide by three: n squared minus seventeen n plus fifty two equals zero. With me? Now factorise. Which two numbers multiply to fifty two and add up to seventeen? Four and thirteen, exactly! So n equals four, or n equals thirteen: two answers, just as I promised. Many students throw one away here. Don't! Let's check both. Four terms: twenty four, twenty one, eighteen, fifteen. They add to seventy eight. Correct! Now keep going. Terms five to thirteen are twelve, nine, six, three, zero, minus three, minus six, minus nine, minus twelve. They cancel in pairs, so they add to zero! So after thirteen terms, the sum is still seventy eight. Mystery solved: both answers are right.

7:14Quick revision: Sn formulas, An^2 + Bn, ratio of kth terms

Clear? So, quick revision. S n is n over two, times two a plus, n minus one, times d, or n over two, times first plus last. The sequence is an A P exactly when S n is capital A n squared plus capital B n, and then d is two times capital A, not capital A itself. For the ratio of k th terms, put n equals two k minus one in the ratio of the sums. Now go and try the practice problems on this page. You can do them. All the best!