JEE MAIN & ADVANCE 11th PCM Maths Demo Videos

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Hello, students, welcome back. In this module we will discuss the Method of Difference. Now, let us see, what does this concept mean. Method of difference, let say I am given sequence a1, a2, a3 till an and this sequence doesn’t look like A.P. or G.P. or A.G.P. and we have to find the sum of this sequence. Now, students, if we are able to find the nth term we can use summation concept. And, students, what you observe over here is, that the difference between consecutive terms is forming an A.P. or G.P. The given sequence is not an A.P. or GP but the difference between the consecutive terms is forming an A.P. or G.P. Then, students, we can use this concept to find the nth term. Students, what I am going to do is, I have written the sum of this series S as the first equation. I am going to rewrite the given sum and I am writing that as second equation. Now, students, I am going to subtract these two series. I know subtracting these two series, S will cancel out with S will obtain zero. And I am going to subtract, students, see over here. S minus S will become equal to zero. a1 the first term remains as it is, a2 minus a1, I have formed in the bracket, a3 minus a2 I have formed second bracket and so on till an minus an minus 1. The last an over here which should be over here and minus an has gone over there on the left hand side. Now, students, an is over here and now, students, we know that difference between the consecutive terms is forming A.P. or G.P. therefore a2 minus a1, a3 minus a2 and so on, can be written in form of A.P. or G.P. and hence their sum could be found out. And hence, I will be able to find out an that is the general term and now since I am able to find the general term everything comes down to this summation series and we know, how to do that.

Now, students, let us take an example and understand exactly what we have to do over here. Let’s say students, I am given the series 5, 7, 11, 17, 25 and so on. And I have to find the sum of nth term of this series. Students, note 5, 7, 11, difference between 7 and 5 is 2, 11 and 7 is 4, 17 and 11 is 6, so difference is 2, 4, 6 and so on. So, difference is forming in A.P. 2, 4, 6 and so on. Therefore students, over here to find the nth term, I will use the method of difference. Sum S, I have already written. Again S I am going to write it but shifting the terms, taking the differences, students, so when I subtract I will get tn. You can see over here, students, on the right hand side I have done this, over there on the left hand side there is zero, tn from here will go on left hand side. So, tn will be equal to 5 plus, you can see over here, students, the bracket there are n minus 1 terms. First term is 2 then 4, 6, 8 that is an A.P. that is sum of n minus 1 terms in A.P. sum of n minus 1 term in A.P. I have written the formula. It just a simple calculation, students. So, tn comes out as n square minus n plus 5. So, when I have to find this sum of series, it’s simple summation of tn and now the sum of terms would be r raised from 1 to n, summation r square minus summation r raised from 1 to n from r plus 5 times summation r raised from 1 to n again 1. Now, it’s the simple application of formula, friends, summation r square is n into n plus 1 into 2n plus 1 divided by 6. Summation r is n into n plus 1 by 2 and we know summation r raised from 1 to n, 1 is simple n, so it becomes 5n. Calculating, is simple calculation, friends, taking out the common and now taking the LCM and then now simplifying it. I obtain the final result, n by 6 into 2n square plus 28. That gives us the final answer, students.

Let us take one more example and understand it. So, let’s say, students, I have to find the sum of series 1, 5, 11, 19, 29, you have to find the sum of these n terms. Now, 1 and 5 difference is 4, 5 and 11 difference is 6, 11 and 19 difference is 8. So, students we can see 4, 6, 8, so the difference again forms an A.P. So again to find the nth term, students, I will use the method of difference. Rewriting the series you can see over here, students, taking the difference on subtracting, I again obtain this series, this equation is obtained taking tn on the left hand side tn will be equal to 1 plus sum of n minus 1 terms. First term is 4, common difference is 2. So, n by 2 into 2n minus n to t. It’s simple calculations, students, calculating it, expanding the brackets to simplify. So, when I expand tn comes out as n square plus n minus 1. Sum of terms that is Sn would be the summation r raise from 1 to n, tr and that would be equal to summation r raise from 1 to n r square plus r minus 1. Now, splitting this summation and now we obtain summation r raise from 1 to n and r square plus summation r raise from 1 to n r minus summation r raise from 1 to n, 1. Substituting the formula, friend, that is very easy, now we have remembered that formula and now its calculation. So, taking out common and then calculating it and so, friends, we calculate and we obtain the final result to it. And final result comes out as that would be equal to n into n square plus 3n minus 1 whole divided by 3.

Students, I hope you understand this example, thank you.

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