How to Score Maximum in SSLC Examination Part 1 | Covid - 19 2020-21 |
Jan 28, 2021 05:03 · 1520 words · 8 minute read
hi my dear students expressionally the mathematician is very important for the life easy maths so on the basis of that one definitely you can go through your own your own goal and definitely your ambitions comes true so i am tamanna so my channel is easy math by the helping of this you can reach your goal so you know that one this year is very difficult here for the students especially education department facing so many problems when compared to other departments students think about their examination the examination may be easy or difficult they don’t know about that one so examination is based on some of the main themes the education department declared that some of the chapters are deleted or not so which chapter is related which syllabus is not there which competency is there which one is important which is which one is not important this type of equations are there in the student’s mind so so the student having a particular way on the basis of this video definitely this 12 20 20 and 21 is a big question mark for the student so this question mark is solved only only only on the base of this video a student so the students are having a special type of a way for going through so expressionally so that some of the chapters are deleted and some of the competencies are also deleted definitely i will go through on that one so first you go through on first starting of this video to at the end of the video definitely you watch then only you can understand which one is there and which one is not the student having two type of roots for the purpose of the facing of examination so if this the student think that one this examination is easy or the examination is difficult is if the examination easy no problem at all because that competency is learnt already by the student in the classroom situations but the examination is difficult then how the examination how is a big problem for you because that type of problems you can’t do very clearly you can’t you can’t get the answer so due to that reason if the examination is difficult then the application level of questions are solved on the base of this video and each and every competencies which is deleted is not explained and which competency which question is repeated in the examination which type of questions are may be coming in your examination all of these things is explained by the helping of this video with the article easy math let’s start now my dear students so on the basis of mathematics syllabus so which chapter is fully there or which is not in the syllabus and which competencies occur and which competency is deleted and which chapter is there so on the basis of first chapter in the tense stand the automatic progression in a state syllabus so this chapter is deleted or not yes yes definitely this chapter is fully occur in this syllabus so arithmetic progression having a main theme to gain so much of so many marks in the examination maybe the weightage of this chapter is around 10 marks is there so there are two topics comes under in atomic arithmetic progression before that one ah i will explain on this session on the basis of so many questions are repeated in the question paper that question papers are in application level the three marks or four marks questions so that question is explained here and definitely you can gain the knowledge and easily you um solve the problem on the base of this video so here two topic is that the the general term another one is the sum of nth term so in arithmetic progression the main two formulas are there a plus n minus 1 into d so here a n is a term that means when which term so when we ask a question according to the term then then we can go through one of this formula that is again in the yes and yes n is uh indicated um the indicated term that is a sn which is comes under sum of nth term so sn is equal to n by 2 into 2 a plus n minus 1 into d so so we can split this one on the base of the multiplication process a into a plus n minus 1 into d yeah so n by 2 so here we can come across this a plus n minus 1 into d is the same as that one a we can write it as a plus a n so one a is there and remaining a is comes under the under in the formula due to that reason i write it up only one a and then another one thing is a n so yes and so here main three formulas occur one n is equal to a plus n minus 1 into d another one is a s n is equal to n by 2 into 2 a plus n minus 1 into d another one is s n is equal to n by 2 into a plus a n so these three formulas are very important so very important to solve so many problems so next we go through on the application level of problem which is repeatedly comes in a question papers so go through on that one so here the question paper repeated question is there first of all we take it as for what value of k will k plus 9 2 k minus 1 2 k plus 7 are consecutive terms of an ap so here you know that when k plus 9 2k play minus 1 and 2k plus 7 is there so these are the three terms given in the ap ap means arithmetic progression so here k plus 9 we we have to take it as a 1 and 2 k minus 1 is take it as a 2 and 2 k plus 7 is take it as a 3 now we get it a 1 a so the formula of a common difference is a2 minus a1 a2 minus a1 and another one is a3 minus a2 so on the base of that one so here given that for what value of k will k plus 9 2 k minus 1 and 2 k plus 7 are consecutive terms of an ap so these are the consecutive term is given in the question per question but so even we don’t know because the k is unknown quantity unknown letter is there due to that reason we don’t know about a one so the common difference is also not given here so due to that reason first of all we find out the common difference so a2 minus a1 having the same value a2 minus a1 is also the same value and a3 minus a2 is also the same value so we equal it and on the base of that we find out the value of k so a 2 minus a 1 is equal to a 3 minus a 2 a 2 minus a 1 is where a 2 is 2 k plus 1 so not plus that one is a minus 2k minus 1 minus is there one is k plus 9 is equal to a 3 is 2 k plus 7 and minus is there put a bracket here a 2 is 2 k minus 1 simplify this one 2k minus 1 so plus into minus minus k plus into minus minus 9 is equal to 2k plus 7 plus into minus minus 2k minus into minus plus 1 so this 2k here 2k and this one is k so dude when we subtract it when we subtract it 2 k minus 1 that is a k minus 10 minus 1 minus 9 is a minus 10 here 2 my minus 2k plus 2k minus 2k get cancelled it obtained 8 k is equal to 8 minus 10 and this is plus 10 so finally k is equal to 18 the value of k is 18 here by the helping of this one we find out the value of k so this type of application level of question some of the questions are a little bit difficult to understand but it is easy when we go through the depth so on the base of this one you find out easily the value of k so only on the base of you know that one a1 which one is a1 which one is d and which type of formula we have to apply here this is a main question which is repeated in a question paper so go through one different problem in the next step next part i explain a more application level question according to the arithmetic progression so the student you enjoy a lot and go through on a different problem thank you.