 Hello and welcome to the session. I am Arsha and I am going to help you with the following question which says the p at q at an r at terms of an ap are a, b and c respectively. Show that q minus r into a plus r minus p into b plus p minus q into c is equal to 0. Let us now begin with the solution. Let the first term of an ap is equal to capital A and the common difference is equal to capital D. So the pth term which is denoted by ap will be equal to a plus p minus 1 into d. And q at term will be aq which is equal to a plus q minus 1 into d and the rth term is equal to ar which is a plus r minus 1 into d. Now we are given that the pth term is a, q at term is b and r at term is c. So let us now take the left hand side of what we have to prove. That is q minus r into a plus r minus p into b plus p minus q into c. So this is equal to q minus r into a is a plus p minus 1 into d plus r minus p into b is a plus q minus 1 into d plus p minus q into a plus r minus 1 into d. So this is further equal to q a plus q into p minus 1 into d minus r a minus r into p minus 1 into d. Now simplifying the next one will give us plus r into a plus r into q minus 1 into d minus p into a minus p into q minus 1 into d plus p into a plus p into r minus 1 into d minus q a minus q into r minus 1 into d. So now q a cancels out with minus q a minus r a with plus r a minus p a with plus p a and so we further have q p minus q d. As simplifying this one we have minus r p plus r into d as simplifying this we have plus r q minus r d minus p q plus p d then we have plus r minus p d minus q r plus q d. Now q p cancels out with minus q p minus q d with plus q d minus r p with plus r p r d with minus r d r q with minus r q and p d with minus p d this is equal to 0 which is the right hand side minus r into a plus r minus p into b plus p minus q into c is equal to 0. So this completes the session take care and have a good day.