Last updated at Dec. 8, 2016 by Teachoo

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Ex 9.2 , 12 The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1). We know that Sn = n/2 ( 2a + (n – 1)d ) Where, Sn = sum of n terms of A.P. n = number of terms a = first term and d = common difference Thus, Sum of n terms = Sn = 𝑛/2(2a + (n – 1)d) And Sum of m terms = Sm = 𝑚/2(2a + (m – 1)d) It is given that, ratio of the sums of m and n terms of an A.P. is m2: n2 (Sum of m terms )/(Sum of n terms) = m2/n2 (Sm )/(Sn ) = m2/n2 (m/2[2a + (m − 1)d])/(n/2[2a + (n − 1)d]) = m2/n2 𝑚/𝑛 ([2a + (m − 1)d])/([2a + (n − 1)d]) = m2/n2 (2a + (m − 1)d)/(2a + (n − 1)d) = n/m × m2/n2 (2a + (m − 1)d)/(2a + (n − 1)d) = m/n We need to show that the ratio of mth and nth term is (2m – 1): (2n – 1). Finding nth and mth terms , We know that an = a + (n – 1)d Where, an = nth term of A.P. n = number of terms a = first term and d = common difference So, nth term = an = a + (n – 1)d Similarly, mth term = am = a + (m – 1)d We need to show that the ratio of mth and nth term is (2m – 1): (2n – 1). i.e. (mth term)/(nth term) = ((2m − 1))/((2n − 1)) (a + (m − 1)d)/(a + (n − 1)d) = ((2m − 1))/((2n − 1)) From (1) (2a + (m − 1)d)/(2a + (n − 1)d) = m/n Replacing m with 2m – 1 and n with 2n – 1 (2a + [(2m − 1) − 1]d)/(2a + [(2n − 1) −1]d) = ((2m − 1))/((2n − 1)) (2a + [2𝑚 − 1 − 1]d)/(2a + [2n − 1 − 1]d) " =" ((2m − 1))/((2n − 1)) (2a + [2𝑚 − 2]d)/(2a + [2n − 2]d) " =" ((2m − 1))/((2n − 1)) (2a + 2(m − 1)d)/(2a +2 (n − 1)d) = ((2m − 1))/((2n − 1)) (2[a + (m − 1)d])/(2[a + (n − 1)d]) = ((2m − 1))/((2n − 1)) (a + (m − 1)d)/(a + (n − 1)d) = ((2m − 1))/((2n − 1)) (mth term of A.P)/(nth term of A.P) = ((2m − 1))/((2n − 1)) Thus, ratio of mth and nth term is (2m – 1): (2n – 1). Hence proved.

Chapter 9 Class 11 Sequences and Series

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .