Ex 8.2, 25 - If a, b, c, d are in GP, show (a2 + b2 + c2) - Ex 8.2

part 2 - Ex 8.2, 25 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series
part 3 - Ex 8.2, 25 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series

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Ex9.3,25 If a, b, c and d are in G.P. show that . (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 We know that a, ar , ar2 , ar3, …. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We need to show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 Taking L.H.S (a2 + b2 + c2) (b2 + c2 + d2) Putting values of b = ar , c = ar2 , d = ar3 (a2 + (ar)2 + (ar2)2) ((ar)2 + (ar2)2 + (ar3)2) = (a2 + a2r2 + a2r4) (a2r2 + a2r2 + a2r6) = ["a2(1 + r2 + r4)" ] ["a2r2(1 + r2 + r4)" ] = ["a2" ] ["a2r2" ]"(1 + r2 + r4)" "(1 + r2 + r4)" = ["a2 a2 r2" ]"(1 + r2 + r4)"2 = ["a4 r2" ]"(1 + r2 + r4)"2 = "a4 r2(1 + r2 + r4)"2 Taking R.H.S (ab + bc + cd)2 Putting values of b = ar , c = ar2 , d = ar3 = ( a × ar + ar × ar2 + ar2 × ar3) 2 = ( a2r + a2r3 + a2r5 )2 = ["a2r (1 + r2 + r4)" ]^2 = (a2r)2 (1 + r2 + r4)2 = a4r2 (1 + r2 + r4)2 = L.H.S Thus L.H.S = R.H.S Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo