If x, y and z are in continued proportion, prove that :
(x+y)2(y+z)2=xz\dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z}(y+z)2(x+y)2=zx.
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Given, x, y and z are in continued proportion
∴xy=yz\therefore \dfrac{x}{y} = \dfrac{y}{z}∴yx=zy
⇒ y2 = xz
Taking LHS,
(x+y)2(y+z)2=x2+y2+2xyy2+z2+2yz\dfrac{(x + y)^2}{(y + z)^2} \\[1em] = \dfrac{x^2 + y^2 + 2xy}{y^2 + z^2 + 2yz}(y+z)2(x+y)2=y2+z2+2yzx2+y2+2xy
Substituting y2 = xz in above we get,
⇒x2+xz+2xyxz+z2+2yz⇒x(x+z+2y)z(x+z+2y)⇒xz= RHS \Rightarrow \dfrac{x^2 + xz + 2xy}{xz + z^2 + 2yz} \\[1em] \Rightarrow \dfrac{x(x + z + 2y)}{z(x + z + 2y)} \\[1em] \Rightarrow \dfrac{x}{z} = \text{ RHS }⇒xz+z2+2yzx2+xz+2xy⇒z(x+z+2y)x(x+z+2y)⇒zx= RHS
Hence, proved that (x+y)2(y+z)2=xz\dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z}(y+z)2(x+y)2=zx.
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If 7x - 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of :
(i) 9x+5y9x−5y\dfrac{9x + 5y}{9x - 5y}9x−5y9x+5y
(ii) 3x2+2y23x2−2y2\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2}3x2−2y23x2+2y2
If 4m+3n4m−3n=74\dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4}4m−3n4m+3n=47, use properties of proportion to find :
(i) m : n
(ii) 2m2−11n22m2+11n2\dfrac{2m^2 - 11n^2}{2m^2 + 11n^2}2m2+11n22m2−11n2
Given, x = a2+b2+a2−b2a2+b2−a2−b2\dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}a2+b2−a2−b2a2+b2+a2−b2.
Use componendo and dividendo to prove that :
b2 = 2a2xx2+1\dfrac{2a^2x}{x^2 + 1}x2+12a2x
If x2+y2x2−y2=218\dfrac{x^2 + y^2}{x^2 - y^2} = 2\dfrac{1}{8}x2−y2x2+y2=281, find :
(i) xy\dfrac{x}{y}yx
(ii) x3+y3x3−y3\dfrac{x^3 + y^3}{x^3 - y^3}x3−y3x3+y3
