 Hello and welcome to the session. In this session we discussed the following question which says the sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find AP. We know that the nth term of an AP is given by AN is equal to A plus N minus 1 whole into D where this AN is the nth term of an AP. A is the first term, D is the common difference and N is the number of terms with the key idea to be used in this question. Let's move on to the solution now. We are given that the sum of the 4th and the 8th terms of an AP is 24 that is we have A4 plus A8 is equal to 24 and sum of 6th and 10th terms is 44. So A6 plus A10 is equal to 44. Now A4 is equal to A plus, now in place of N we put 4 minus 1 this whole into D that is we have A4 is equal to A plus 3D. Let this be equation 1. Now A8 is equal to A plus 8 minus 1 whole into D that is we have A8 is equal to A plus 7D. Let this be equation 2. Then A6 is equal to A plus 6 minus 1 whole into D which gives us A6 is equal to A plus 5D. Let this be equation 3. Then A10 is equal to A plus 10 minus 1 whole into D which is A10 is equal to A plus 9D. Let this be equation 4. Now we take let this equation be capital A and this equation be capital B. Now from capital A we have A4 plus A8 is equal to 24. Now substituting the values for A4 and A8 from equations 1 and 2 we get A plus 3D plus A plus 7D is equal to 24 that is 2A plus 10D is equal to 24 which means we have A plus 5D is equal to 12. Let this be equation 5. Now next we have from equation capital B we have A6 plus A10 is equal to 44. Now substituting the values for A6 and A10 from equations 3 and 4 we get A plus 5D plus A plus 9D is equal to 44. Further we get 2A plus 14D is equal to 44. This gives us A plus 7D is equal to 22. Let this be equation 6. Now consider the two equations A plus 5D equal to 12 and A plus 7D equal to 22. Now we will solve both these equations for A and D. So we will subtract these two equations. So this gives us minus 2D is equal to minus 10 that is we get D is equal to 10 upon 2 to 5 times as 10 and therefore we get D is equal to now substituting D equal to 5 in equation 5 that is this equation A plus 5D equal to 12. A plus 5D that is 5 into 5 is equal to 12. A plus 25 is equal to 12 or you have A is equal to 12 minus 25. This gives us A is equal to minus 13. Thus we get A equal to minus 13 and D equal to 5 that is the first term of the AP is equal to minus 13 and the common difference of the AP is equal to 5. Now the AP is given by A A plus D A plus 2D A plus 3D and so on. Thus we have AP is minus 13, minus 13 plus 5, minus 13 plus 10, minus 13 plus 15 and so on that is we have AP is minus 13, minus 8, minus 3, 2 and so on. Thus the required AP is minus 13, minus 8, minus 3, 2 and so on. So this is our final answer. This completes the session. Hope you have understood the solution of this question.