If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1. I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given! S(m)÷s(n) = m²/n² On solving till a point, we get this equation 2an + mnd -nd = 2am + mnd -dm → 2an-nd = 2am-dm → n(2a-d) = m(2a-d) → n=m. ......(1) In the 'to prove' statement, LHS → a+(m-1)d / a+(n-1)d Replacing n with m (using 1) a+(m-1)d/a+(m-1)d = 1/1 = 1 RHS → 2m-1/2n-1 Replacing n with m (using 1) 2m-1/2m-1 = 1/1 =1 Hence, LHS=RHS. IS THIS CORRECT?

If the ratio of sum of m terms and n terms of an A.P. is , m²:n² the

1.	If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1. I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given! S(m)÷s(n) = m²/n² On solving till a point, we get this equation 2an + mnd -nd = 2am + mnd -dm → 2an-nd = 2am-dm → n(2a-d) = m(2a-d) → n=m. ......(1) In the 'to prove' statement, LHS → a+(m-1)d / a+(n-1)d Replacing n with m (using 1) a+(m-1)d/a+(m-1)d = 1/1 = 1 RHS → 2m-1/2n-1 Replacing n with m (using 1) 2m-1/2m-1 = 1/1 =1 Hence, LHS=RHS. IS THIS CORRECT?
Answer» If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1. I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given! S(m)÷s(n) = m²/n² On solving till a point, we get this equation 2an + mnd -nd = 2am + mnd -dm → 2an-nd = 2am-dm → n(2a-d) = m(2a-d) → n=m. ......(1) In the 'to prove' statement, LHS → a+(m-1)d / a+(n-1)d Replacing n with m (using 1) a+(m-1)d/a+(m-1)d = 1/1 = 1 RHS → 2m-1/2n-1 Replacing n with m (using 1) 2m-1/2m-1 = 1/1 =1 Hence, LHS=RHS. IS THIS CORRECT?