 Hello and welcome to the session. I am Asha and I shall be helping you with the following question that says the ratio of the sums of m and n terms of an AP is m square is to n square show that the ratio of mth term and nth term is 2m minus 1 is to 2n minus 1. So let us now begin with the solution and let A be the first term, D be the common difference of AP series then we are given that the sum of m terms. So first let us denote sum of n terms is equal to Sn. Now we are given that sum of m terms upon sum of n terms is equal to m square upon n square. Now the formula of Sn is m upon 2 into 2A plus m minus 1 into D upon n upon 2 into 2A plus n minus 1 into D is equal to m square upon n square or we have 2A plus m minus 1 into D upon 2A plus n minus 1 into D is equal to m square upon n square into n upon m. Now n cancels out with n and m with 1m so we have m upon n here 2 cancels out with 2 therefore on cross multiplying n goes the numerator of the right hand side and m on the denominator. So we have 2A plus m minus 1 into D upon 2A plus n minus 1 into D is equal to m upon n. Let this be equation number 1. Now we have to find the ratio of mth term is to nth term and An upon An is equal to A plus m minus 1 into D upon A plus n minus 1 into D. Now on multiplying the numerator and denominator by 2 this can further be written as mth term upon nth term equal to 2A plus 2 times of m minus 1 into D upon 2A plus 2 times of n minus 1 into D. Let this be equation number 2. Now L it is of 1 which is 2A plus m minus 1 into D upon 2A plus n minus 1 into D will be identical with 2 right hand side of 2 if m minus 1 replaces 2n minus 2 and n minus 1 replaces 2 times of n minus 1. So replacing m by 2n minus 1 and n by 2n minus 1 on both sides of 1 we will get the arches of equation 2. So on replacing minus 1 and n by minus 1 on both sides equation 1 we get minus 2 into D upon 2A plus 2n minus 2 into D. This is equal to 2m minus 1 upon 2n minus 1 and this is 2A plus 2 times of m minus 1 into D upon 2A plus 2 times of n minus 1 into D which is equal to 2m minus 1 upon 2n minus 1. So this is the mth term and this is the nth term. So we have 2m minus 1 upon 2n minus 1 and this shows that the ratio of mth nth term is equal to 2m minus 1 is to 2n minus 1. So this completes the session. Take care and have a good day.