 Hello and welcome to the session. In this session we discuss the following question which says find four numbers in AP whose sum is 20 and the sum of whose squares is 200. So let's see how we can find the four numbers with sum as 20 and sum of the squares of these four numbers is given as 200. So we assume let the numbers be A minus 3D, A minus D, A plus D and A plus 3D. Now it's given that the numbers A minus 3D, A minus D, A plus D and A plus 3D are in AP. Now the sum of these numbers that is A minus 3D plus A minus D plus A plus D plus A plus 3D is given as 20. So from here we get A minus 3D plus A minus D plus A plus D plus A plus 3D is equal to 20. Now minus 3D and 3D gets cancelled, minus D and D gets cancelled. So from here we get 4A is equal to 20. This gives us A is equal to 20 upon 4. Now 4, 5 times is 20. Therefore we get A is equal to 5. It's also given that the sum of these squares of these four numbers is 200. So we have A minus 3D whole square plus A minus D whole square plus A plus D whole square plus A plus 3D whole square is equal to 200. Now we open these brackets using the identities. So we get A square minus 6AD plus 3D whole square plus A square minus 2AD plus D square plus A square plus 2AD plus D square plus A square plus 6AD plus 3D square is equal to 200. So this gives us A square minus 6AD plus 9D square plus A square minus 2AD plus D square plus A square plus 2AD plus D square plus A square plus 6AD plus 9D square is equal to 200. Now this 6AD and minus 6AD cancels, 2AD minus 2AD gets cancelled. Then we have 4A square plus 20D square is equal to 200. Now taking 4 common from here, so we get inside the bracket A square plus 5D square and this is equal to 200. So from here we get A square plus 5D square is equal to 200 upon 4. Now 54 times is 200. So we get A square plus 5D square is equal to 50. Now substituting the value of A equal to 5 that we have found out, we get 5 square plus 5D square is equal to 50. That is, we have 25 plus 5D square is equal to 50. From here we get 5D square is equal to 50 minus 25 that is 5D square is equal to 25. So this gives us D square is equal to 25 upon 5 which is equal to 5. Thus we have D is equal to plus minus 5. So now we have got A is equal to 5 and D is equal to plus minus 5 and the given numbers were A minus 3D, A minus D, A plus D and A plus 3D. Now when we have D is equal to 5 then the numbers are given by substituting A equal to 5 and D equal to 5 in these numbers. So this would give us 5 minus 3 into 5 then the second number would be 5 minus 5. The next number would be 5 plus 5 then the next would be 5 plus 3 into 5 that is we have 5 minus 15 that would be minus 10, 5 minus 5, 0, 5 plus 5, 10, 5 plus 15 would be 20. So thus we say when D is equal to 5 the numbers minus 10, 0, 10, 20 are in AP. Now we take the other value of D that is minus 5 so when D is equal to minus 5 the numbers would be given by 5 minus 3 into minus 5 then 5 minus of minus 5, 5 plus of minus 5 then 5 plus 3 into minus 5. That is the numbers will be given by 5 plus 15 that is 20, 5 plus 5, 10, 5 minus 5, 0 and 5 minus 15 would be minus 10. Thus when D is equal to minus 5 the numbers 20, 10, 0, minus 10 are in AP. Thus the required numbers would be 20, 10, 0, minus 10 or minus 10, 0, 10, 20. So this completes the session. Hope you have understood the solutions for this question.