 Hello friends welcome to the problem-solving session on selection of terms of an AP in the given problem It's Given that find four numbers in AP whose sum is 20 and some of whose squares is 120 Now they're asking to find four numbers in AP right whose sum is 120 if you recollect in the previous session We talked about how You know the selection of those terms in a particular format is going to help us so in this case I can always assume that the terms four terms four terms in AP Be what all? a minus 3d a minus d a Plus d and a plus 3d now you can ask me. Why can't we take? Simply a a plus d a plus 2d and a plus 3d and my answer would be only You know to bring in some kind of comfort while solving this problem You can always do this, but then what will happen is you will have to play with Equations with two variables. So hence we are trying to you know Play it a little smartly and then we are eliminating the variable because we know that APs in APs the consecutive Terms have constant difference So we are exploiting that particular property to so that we deal with only one variable and Especially when some of the terms is given Okay, it will become very clear in the next step. So some of some is 20 some of all those terms So you can see a minus 3d Plus a minus d these are the terms and I'm adding them and it is given that some of all these terms is Is is 20 so let it be like this 20, right? So now you will see the advantage of taking it as a minus 3d a minus d a plus d and a plus 3d if you look closely all these terms Sorry the D terms Will get cancelled. Okay, right leaving only for a behind and this is 20 So immediately you get a value of a which is 5 correct, so you are You know you eliminated one variable and now you know one variable at least that is a it will be very easy from the second Relationship to find out D and hence You will be able to find out the entire AP or some of school whose choir. So what is the square of these terms? a minus 3d Whole square plus a minus d whole square Plus a plus d whole square Plus a plus 3d whole square is it and this is given to be equal to 120 Okay, now let's try in solve this so Opening the brackets, you'll get a square minus 6 ad plus 9d square Plus a square. This is from the first expansion second expansion a square minus 280 plus d square second expansion Similarly third expansion will be a square plus 2 ad plus d square and Fourth one is a square plus 6 ad plus 9d Square right and this entire thing is 120 now you can see 6 ad minus 6 ad will go To ad and plus 280 will go. So what's left for a square? So we are left with four a squares and then 20 d square and This is 120 and thankfully we have the value of a here five. So we can write this as four into five square is 20 d square sorry plus 20 d square is 120 clear so hence 20 d square is 120 minus four into five square is hundred So this is 20. So what do I get d square is one? Correct. That means d is plus or minus one Okay, so now we have got both a a we got this this one is a and We got d Okay, so now we can find out the AP right or the four terms. So let me try it here. So the four terms are First term was a minus 3d that means it could be five plus one or Five minus one that is six or four am I right? Five oh minus 3d. I'm sorry. No, this is three and yeah, so you have to just a correction guys This is five plus three and five minus three. So hence either it is eight or two, right? This is not a division sign. So let me remove this Either or this so let me put a line like that. So there are two two First terms eight or two next is a minus d a minus d will be simply five plus one that is six here Five minus one that is four then a plus d will be now you can guess this is four and And This one is Six and then a plus 3d is Eight no, sorry two in this case. So I'm taking d as minus one and then plus one Okay, and then this one will be eight So the two APs or two four terms in AP are eight six four two or two four six eight both are Correct solution. Okay. So hence you learned in this problem that You know if you take the four terms as a minus 3d a minus d a plus d and a plus 3d Then it helps in eliminating one of the variables and hence the problem solving becomes a little easier