 Hello and welcome to this session. In this session we discuss the following question which says, if A, B, C and D are in continued proportion, proves that AQ plus BQ plus CQ upon BQ plus CQ plus DQ is equal to A upon D. We know that when three quantities say A, B and C are in continued proportion, then A upon B is equal to B upon C. This is the key idea that we use for this question. Let's proceed with the solution now. We are given that A, B, C and D are in continued proportion. Therefore, we can say that A upon B is equal to B upon C is equal to C upon D. And we are supposed to prove that AQ plus BQ plus CQ and this whole upon BQ plus CQ plus DQ is equal to A upon D. So, first of all, we take A upon B is equal to B upon C is equal to C upon D is equal to K. This means that A is equal to BK, B is equal to CK and C is equal to DK. Or A would be equal to BK and in place of B we put CK. So, CK into K that is CK square and in place of C we put DK. So, it would be DK into K square that is equal to DKQ. Therefore, we have A is equal to DKQ. Now, consider B equal to CK in place of C we put DK. So, DK into K is equal to DK square. Therefore, B is equal to DK square. Now, consider the LHS which is equal to AQ plus BQ plus CQ upon BQ plus CQ plus DQ. Now, here in place of A we put DKQ and in place of B we put DK square and in place of C we put DK. So, we have this is equal to DKQ the whole cube plus DK square the whole cube plus DK the whole cube. As we have substituted the values for AB and C and this whole upon DK square the whole cube plus DK the whole cube plus DQ. So, we get this is equal to DQ into K to the power 9 plus DQ into K to the power 6 plus DQ into K cube. This whole upon D cube into K to the power 6 plus D cube into K cube plus D cube. Now, taking D cube common in the numerator we have D cube into K to the power 9 plus K to the power 6 plus K to the power 3. And this whole upon taking D cube common in the denominator we have D cube into K to the power 6 plus K to the power 3 plus 1 the whole. Now, this D cube D cube cancels. Now, again we can take K cube common from the numerator. So, here we have K cube into K to the power 6 plus K to the power 3 plus 1 the whole and this whole upon K to the power 6 plus K cube plus 1. Now, these two expressions cancels and this is equal to K cube that is we have LHS is equal to K cube. Now, we consider the RHS which is equal to A upon D. Now, in place of A we put D K cube upon D. Now, D and D cancels and this is equal to K cube. Thus we have the RHS is equal to K cube. So, LHS and RHS each is equal to K cube. So, we say the LHS is equal to the RHS. Hence, we have proved that A cube plus B cube plus C cube and this whole upon B cube plus C cube plus D cube is equal to A upon D. This is what we were supposed to prove. So, this completes the session. Hope you have understood the solution of this question.