JEE MAIN & ADVANCE 11th PCM Maths -Sequence & Series-2 Demo Videos

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Hello students welcome back in this model will discuss the method of difference now let’s see what does this concept means on method of difference let’s as students I’m given a sequence a1 a2 a3till  an and this sequence doesn’t looks like an AP or GP or AGP and we have to find the sum of these sequence now students as we know that if we are able to find the nth term  we can use this summation concept understands what you observe where is that the difference between consecutive terms is forming an ap or GP the given sequence is not ap or GP but the difference between the conjugative  terms  is forming an AP or GP then  we can use this concept to find the nth term  students what I’m going to do is I written the solve the series as the first equation i’m going to rewrite the given sum and i’m writing that as the second equation now students i’m going to subtract these two series I know subtracting these two series S will cancel out with this will obtain 0 when i’m going to subtract this one see over here s – s will become equal to 0 a1 The first  term remains as it is a2 minus a1 have formed in the bracket a3 and a4 from the second bracket and so on till an minus an minus 1 the last an over here which was over here should be minus an has gone over there on the left-hand side Now students so an is over here and now students we know that difference between the conjugative  terms is forming an AP or GP there for a 2- a1,a3 minus a2 and so on can be written in form of Ap GP and hands there some could be found out and hence i’ll be able to find an that is the general term and now since i’m able to find a 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 what exactly want to do over here now students I’m given the series that is 5 ,7 ,11, 17 and so on and I want to find some of in terms of the series in students note five seven eleven difference between seven and five is two 11 – 7 is 4, 17 – 11 is 6 so differences 2, 4,6 and so on so difference is forming an AP 2,4,6 and so on therefore his students over here to find the nth term I’ll use the method of difference sum s I have already written on again sn unknown but shifting the terms taking the differences students when i subtract i will get tn and you can see over his students on the right hand side I have done this or there on the left-hand side there is 0 tn from here we’ll go on the left hand side so TN will be equal to 5 plus you can see where here students the bracket there are n minus 1 terms 1st term is 2 then 4,6,8 that is an AP that is sum of n minus 1 terms in AP sum of n minus 1 terms an AP write down the formula it’s just simple calculation now students Tn and terms are n square minus n plus 5 so when I were to find the sum of series it’s simple summation of tn and now the sum of terms will be varies from 1 to n summation r square minus summation r varies from 1 to n R plus 5 terms the summation from 1 to n and again one now it’s a simple application of formulas students summation r square into n plus 1 into 2n plus 1/6 summation r is n into n plus 1 by 2 and we know summation on 1 to n  and one is simple and so it becomes 5n calculating is simple calculation is students taking out the common and now taking the LCM and then now simplifying it .I obtained  the final result n by 6 into 2 n square plus 28 that give us the final answers students let’s take one more example under stranded so let’s different i will find some of the series 1,5 ,11 ,19 ,29 where to find the sum of these nth terms now one in five differences 4, 5 11 differences 6, 11 and  19 differences 8 so as students  we can see 4,6,8 the difference again forms an AP so again to find the nth term i use the method of difference rewriting the series can see over us students taking the difference on subtracting i will obtained  up in 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 term   1st term  is 4 common differences too so n/2  2a plus and minus 1 *T simple calculation now students calculating it expanding the bracket to simplify so when I expand tn comes out so n square plus n minus 1, sum of terms that is SN will be summation  are varies from 1 to n TR and that would be equal to some Summation  at from 1 to n R squared plus R minus 1 now splitting the summation  and we obtained  summation r very someone to an R squared plus summation  varies from one to n are  summation R varies from 1 to N 1 substituting the formulas  students that very easy now we have to remember that formula and now its calculation so taking out common and then calculating it to his students we calculate and we obtain the final result and final result comes out as that would be equal to n into n square plus 3 n minus 1 whole divided by 3 students i hope you understand this Example thank you

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