Arithmetic Progression

Chapter 8 Class 11 Sequences and Series
Serial order wise

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### Transcript

Ex 9.2 , 15 If (π^π + π^π)/(π^(πβ1) + π^(πβ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 (π^π + π^π)/(π^(πβ1) + π^(πβ1) ) So, (π^π + π^π)/(π^(πβ1) + π^(πβ1) ) = (a + b)/2 2(an + bn) = (a + b) (an β 1 + bn β 1) 2an + 2bn = a(an β 1 + bn β 1) + b(an β 1 + bn β 1) 2an + 2bn = aan β 1 + abn β 1 + ban β 1 + bbn β 1 2an + 2bn = a1 . an β 1 + abn β 1 + ban β 1 + b1 . bn β 1 2an + 2bn = a1 + n β 1 + abn β 1 + ban β 1 + b1 + n β 1 2an + 2bn = a1 + n β 1 + abn β 1 + ban β 1 + b1 + n β 1 2an + 2bn = an + abn β 1 + ban β 1 + bn 2an + 2bn β an β abn β 1 β an β 1 b β bn = 0 2an β an + 2bn β bn - abn β 1 β an - 1 b = 0 an + bn β abn β 1 β an β 1 b = 0 an β an β 1 b + bn β a bn β 1 = 0 a.an β 1 β an β 1 b + b.bn β 1 β a bn β 1 = 0 an β 1 (a β b) β bn β 1 (a β b) = 0 (an β 1 β bn β 1)(a β b) = 0 β΄ an β 1 β bn β 1 = 0 Solving an β 1 = bn β 1 an β 1 = bn β 1 π^(π β1)/(π^(π β1) ) = 1 (π/π)^(π β1) = 1 (π/π)^(π β1) = (π/π)^0 Comparing powers n β 1 = 0 n = 1 Hence n = 1