 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says if A, B and C are in AP, B, C and D are in GP and 1 upon C or 1 upon D and 1 upon A are in AP prove that A, C and E are in GP. Let us now start with the solution. And here we are given first that A, B, C are in AP. So this implies B minus A is equal to C minus B or you can say that 2 times B is equal to A plus C. Also we are given that B, C and D are in GP. This implies C upon B is equal to D upon C or we can say that C square is equal to B, D. And last we are given that 1 upon C, 1 upon D and 1 upon E are in AP. So this again implies that 2 times of 1 upon D is equal to 1 upon C plus 1 upon E which can further recognize D is equal to 2 times of CE upon C plus E. We have to show that is we have to show that C square is equal to AE. Let us start with C square. The C square generated as B into D this is from C square is equal to B is A plus C upon 2 from 1 and D is 2 CE upon C plus E. Now on cross multiplying we have 2 times into C square is equal to 2A CE plus 2C square E which further implies 2C cube plus 2EC square is equal to 2A CE plus 2C square E. Now the query cancels up with 2C square E and we have C cube minus 2A CE is equal to 0 which further implies 2 times of C into C square minus taking 2 and C common we have A left. So this is equal to 0. Now if the product of 2 numbers is equal to 0 then either one of them is 0. So this implies either 2C is equal to 0 or C square minus A is equal to 0. But C cannot be 0 since we are given that 1 upon C, 1 upon D and 1 upon E are in AP. 1 upon C if C is equal to 0 then 1 upon C is equal to infinity thus 2C is not equal to 0 and this further implies that C square minus AE is equal to 0 or C square is equal to AE or we can say that C upon A is equal to E upon C or we have A and say that A, C and E are in risk of each cessation take care and have a good day.