Ex 8.2, 22 - If pth, qth and rth terms of a GP are a, b, c - Ex 8.2

part 2 - Ex 8.2, 22 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series
part 3 - Ex 8.2, 22 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series
part 4 - Ex 8.2, 22 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series
part 5 - Ex 8.2, 22 - Ex 8.2 - Serial order wise - Chapter 8 Class 11 Sequences and Series

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Ex9.3, 22 If the pth ,qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq r br p cp q = 1 We know that nth term of G.P = ARn 1 (We are using a, r in the question, so we use A for first term and R for common ratio) It is given that pth term of G.P = a Ap = a ARp 1 = a a = ARp 1 aq r = ("ARp 1")q r We need to show that aq r br p cp q = 1 Also, qth term of G.P = b Aq = b ARq 1 = b b = ARq 1 br p = (ARq 1)r p & rth term of G.P = c Ar = c ARr 1 = c c = ARr 1 cp q = (ARr 1)p q Now, our equations are aq r = ("ARp 1")q r (1) br p = (ARq 1)r p (2) & cp q = (ARr 1)p q (3) Taking L.H.S. aq r br p cp q Putting values from (1), (2) & (3) = ("ARp 1")q r (ARq 1)r p (ARr 1)p q = "Aq r R(p 1")(q r) "Ar p R(q 1")(r p) "Ap q R(r 1")(p q) = "Aq r Ar p Ap q " "R(p 1")(q r) "R(q 1")(r p) "R(r 1")(p q) = A(q r) + (r p) + (p q) R(p 1) (q - r) + (q 1) ( r - p) + (r 1) ( p-q) = A(q r) + (r p) + (p q) Rp(q r ) 1(q r) + q( r p) 1(r p) +r(p q) 1( p q) = A(q q) + (r r) + (p p) Rpq pr q + r + qr qp r + p + rp qr p + q = A0 R pq pq q + q + r r + p p + qr qr + qp qr = A0 R0 = 1 1 = 1 = R.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