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Ex 9.2, 15 - If an + bn / an-1 + bn-1 is AM between a, b - Ex 9.2

Ex 9.2, 15 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 15 - Chapter 9 Class 11 Sequences and Series - Part 3

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Ex 9.2 , 15 If (𝑎^𝑛 + 𝑏^𝑛)/(𝑎^(𝑛−1) + 𝑏^(𝑛−1) ) is the A.M. between a and b, then find the value of n. We know that arithmetic mean between a & b is A.M. = (a + b)/2 It is given that AM between a & b is (𝑎^𝑛 + 𝑏^𝑛)/(𝑎^(𝑛−1) + 𝑏^(𝑛−1) ) So, (𝑎^𝑛 + 𝑏^𝑛)/(𝑎^(𝑛−1) + 𝑏^(𝑛−1) ) = (a + b)/2 2(an + bn) = (a + b) (an – 1 + bn – 1) 2an + 2bn = a(an – 1 + bn – 1) + b(an – 1 + bn – 1) 2an + 2bn = aan – 1 + abn – 1 + ban – 1 + bbn – 1 2an + 2bn = a1 . an – 1 + abn – 1 + ban – 1 + b1 . bn – 1 2an + 2bn = a1 + n – 1 + abn – 1 + ban – 1 + b1 + n – 1 2an + 2bn = a1 + n – 1 + abn – 1 + ban – 1 + b1 + n – 1 2an + 2bn = an + abn – 1 + ban – 1 + bn 2an + 2bn – an – abn – 1 – an – 1 b – bn = 0 2an – an + 2bn – bn - abn – 1 – an - 1 b = 0 an + bn – abn – 1 – an – 1 b = 0 an – an – 1 b + bn – a bn – 1 = 0 a.an – 1 – an – 1 b + b.bn – 1 – a bn – 1 = 0 an – 1 (a – b) – bn – 1 (a – b) = 0 (an – 1 – bn – 1)(a – b) = 0 ∴ an – 1 – bn – 1 = 0 Solving an – 1 = bn – 1 an – 1 = bn – 1 𝑎^(𝑛 −1)/(𝑏^(𝑛 −1) ) = 1 (𝑎/𝑏)^(𝑛 −1) = 1 (𝑎/𝑏)^(𝑛 −1) = (𝑎/𝑏)^0 Comparing powers n – 1 = 0 n = 1 Hence n = 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.