Ex 9.3, 22 - Class 11

Transcript

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