 Hello and welcome to the session. I'm Asha and I shall be helping you with the following question which says, sum of first P, Q and R terms of an APR, ABC respectively, prove that A upon P into Q minus R plus B upon Q into Q R minus P plus C upon R into P minus Q is equal to C low. Let us now begin with the solution and here we are given that the sum of first P terms is equal to A. The formula of sum of first P terms is P upon 2 into 2A1 where A1 is the first term of the AP series plus P minus 1 into D where D is the common difference. So here A1 is equal to the first term and D is the common difference. So let this be equation number one. Now we are also given that sum of first Q terms is equal to B, so that is Q upon 2 into 2A1 plus Q minus 1 into D is equal to B and lastly sum of first R terms is equal to C, so this implies R upon 2 into 2A1 plus R minus 1 into D is equal to C. So let this be equation number two and this be equation number three. Now one can further be written as 2A1 plus P minus 1 into D upon 2 is equal to A upon P. So let this be equation number four. Two can be written as 2A1 plus Q minus 1 into D upon 2 is equal to B upon Q. Let this be equation number five and three can be written as 2A1 plus R minus 1 into D upon 2 is equal to C upon R. So let this be equation number six. Now multiplying equation four by Q minus R by R minus P equation six by P minus Q we have A upon P into Q minus R is equal to Q minus R into A1 plus P minus 1 into D upon 2. So let this be equation number seven. Equation five on multiplying with R minus P can be written as B upon Q into R minus P is equal to R minus P into A1 plus Q minus 1 into D upon 2. This is equation number eight. Equation six can be written as C upon R into P minus Q is equal to P minus Q into A1 plus R minus 1 upon 2 into D. So let this be equation number nine and now adding equation seven, eight and nine we get let's insight A upon P into Q minus R plus B upon Q into R minus P plus C upon R into P minus Q is equal to Q minus R into A1 plus P minus 1 into D upon 2 plus R minus P into A1 plus Q minus 1 into D upon 2 plus P minus Q into A1 plus R minus 1 into D upon 2 which is further equal to taking A1 common we have Q minus R plus R minus P plus P minus Q from these three brackets plus D upon 2 taking common we have Q minus R into P minus 1 plus R minus P into Q minus 1 plus P minus Q into R minus 1 equal to A1 plus Q minus Q cancels out minus R with plus R and minus P with plus P we have 0 plus D upon 2 and here on multiplying we have Q P minus Q minus R P plus R now from here we have plus R Q minus R minus P Q plus P plus now solve it these two brackets we have P R minus P minus Q R plus Q now on cancelling minus Q cancels out with plus Q P Q with minus P Q minus R P with plus R P R with minus R R Q with minus R Q and plus P with minus P so we have 0 plus D upon 2 into 0 so this is again equal to 0 plus 0 which is equal to 0 therefore A upon P into Q minus R plus B upon Q into R minus P plus C upon R into P minus Q is equal to 0 so this completes the session take care and have a good day.