 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says show that the sum of m map plus nth term and m minus nth terms of an AP is equal to the twice of mth term. Let us now begin with the solution and let first term of an AP is equal to A common difference is equal to D and we have to show that sum of m plus nth term is denoted by A m plus n plus its sum of m minus nth term is equal to twice the mth term so two times of AM. Now the m plus nth term is A plus minus 1 into D and A m minus nth term is equal to A plus m minus n minus 1 D and the mth term is A plus D. Now start with the left hand side both we have to show which is A m plus n plus m minus nth term this is equal to A plus m plus n minus 1 into D plus A plus m minus n minus 1 into D which is further equal to 2A plus mD plus mD minus D plus mD minus mD minus D. Now plus mD cancels up with minus mD and we have 2A plus 2 mD minus 2D or we have two times of A plus mD minus D or we have two times A plus m minus 1 into D and this is the mth term of the AP so this is equal to two times of AM and thus we have A m minus n that is the sum of m plus nth term and m minus nth term is equal to twice the mth term is equal to the mth term so this completes the session take care and bye for now.