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Last updated at May 29, 2018 by Teachoo

Misc 13 If , (a+bx)/(aβbx) = (b+cx)/(bβcx) = (c+dx)/(aβdx) (x β 0)then show that a, b, c and d are in G.P. Introduction Componendo dividendo If π₯/π¦ = π/π Applying componendo dividendo (π₯ + π¦)/(π₯ β π¦) = (π + π)/(π β π) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 β 2) = (4 + 8)/(4 β 8) 3/(β1) = 12/(β4) -3 = -3 Misc 13 If , (a+bx)/(aβbx) = (b+cx)/(bβcx) = (c+dx)/(aβdx) (x β 0)then show that a, b, c and d are in G.P. We have (a+bx)/(aβbx) = (b+cx)/(bβcx) = (c+dx)/(c β dx) & we want to show that a, b, c, d are in G.P. Taking (a+bx)/(aβbx) = (b+cx)/(bβcx) = (c+dx)/(c β dx) Applying componendo dividendo (a + bx + a β bx)/((a + bx) β(aβbx)) = (b + cx + (b β cx))/(b + cx β(b β cx)) = (c + dx + (c β dx))/(c + dx β (c β dx)) (a + a + bx β bx)/(ππ₯+ bx β a + a ) = (b + b + cx β cx)/(cx + cx β π + π) = (c + dx + c β dx)/(dx + dx β π + π) (2π+0)/(2ππ₯+0) = (2π + 0)/(2ππ₯ + 0) = (2π+0)/(2ππ₯+0) 2π/2ππ₯ = 2π/2ππ₯ = 2π/2ππ₯ π/ππ₯ = π/ππ₯ = π/ππ₯ a/b " =" b/c = c/d b/a " =" c/b = d/c Thus, a, b, c & d are in GP because their common ratio is same