Find x, when :
(i) 3 : 4 :: 2.4 : x
(ii) 1 : 3 :: x : 7
(iii) x : 1.5 :: 3 : 5
Answer
(i) Given,
3 : 4 :: 2.4 : x
Solving for x,
⇒ 3 4 = 2.4 x ⇒ 3 × x = 4 × 2.4 ⇒ 3 x = 9.6 ⇒ x = 9.6 3 ⇒ x = 3.2 \Rightarrow \dfrac{3}{4} = \dfrac{2.4}{x} \\[1em] \Rightarrow 3 \times x = 4 \times 2.4 \\[1em] \Rightarrow 3x = 9.6 \\[1em] \Rightarrow x = \dfrac{9.6}{3} \\[1em] \Rightarrow x = 3.2 ⇒ 4 3 = x 2.4 ⇒ 3 × x = 4 × 2.4 ⇒ 3 x = 9.6 ⇒ x = 3 9.6 ⇒ x = 3.2
Hence, x = 3.2
(ii) Given,
1 : 3 :: x : 7
Solving for x,
⇒ 1 3 = x 7 ⇒ x = 7 3 = 2 1 3 . \Rightarrow \dfrac{1}{3} = \dfrac{x}{7} \\[1em] \Rightarrow x = \dfrac{7}{3} = 2\dfrac{1}{3}. ⇒ 3 1 = 7 x ⇒ x = 3 7 = 2 3 1 .
Hence, x = 2 1 3 2\dfrac{1}{3} 2 3 1
(iii) Given,
x : 1.5 :: 3 : 5
Solving for x,
⇒ x 1.5 = 3 5 ⇒ 5 × x = 1.5 × 3 ⇒ x = 4.5 5 ⇒ x = 0.9 \Rightarrow \dfrac{x}{1.5} = \dfrac{3}{5} \\[1em] \Rightarrow 5 \times x = 1.5 \times 3 \\[1em] \Rightarrow x = \dfrac{4.5}{5} \\[1em] \Rightarrow x = 0.9 ⇒ 1.5 x = 5 3 ⇒ 5 × x = 1.5 × 3 ⇒ x = 5 4.5 ⇒ x = 0.9
Hence, x = 0.9
Find the fourth proportional to :
(i) 3, 8 and 21
(ii) 1.4, 3.2 and 7
(iii) 1.5, 4.5 and 3.6
(iv) a2 , ab and b2
(v) (a2 − ab + b2 ), (a3 + b3 ) and (a − b)
Answer
(i) Given,
3, 8 and 21
Let the fourth proportional to 3, 8 and 21 be x,
⇒ 3 : 8 = 21 : x
⇒ 3 8 = 21 x ⇒ x = 21 × 8 3 ⇒ x = 168 3 ⇒ x = 56. \Rightarrow \dfrac{3}{8} = \dfrac{21}{x} \\[1em] \Rightarrow x = \dfrac{21 \times 8}{3} \\[1em] \Rightarrow x = \dfrac{168}{3} \\[1em] \Rightarrow x = 56. ⇒ 8 3 = x 21 ⇒ x = 3 21 × 8 ⇒ x = 3 168 ⇒ x = 56.
Hence, the fourth proportional is 56.
(ii) Given,
1.4, 3.2 and 7
Let the fourth proportional to 1.4, 3.2 and 7 be x,
⇒ 1.4 : 3.2 = 7 : x
⇒ 1.4 3.2 = 7 x ⇒ x = 7 × 3.2 1.4 ⇒ x = 22.4 1.4 ⇒ x = 16. \Rightarrow \dfrac{1.4}{3.2} = \dfrac{7}{x} \\[1em] \Rightarrow x = \dfrac{7 \times 3.2}{1.4} \\[1em] \Rightarrow x = \dfrac{22.4}{1.4} \\[1em] \Rightarrow x = 16. ⇒ 3.2 1.4 = x 7 ⇒ x = 1.4 7 × 3.2 ⇒ x = 1.4 22.4 ⇒ x = 16.
Hence, the fourth proportional is 16.
(iii) Given,
1.5, 4.5 and 3.6
Let the fourth proportional to 1.5, 4.5 and 3.6 be x,
⇒ 1.5 : 4.5 = 3.6 : x
⇒ 1.5 4.5 = 3.6 x ⇒ x = 4.5 × 3.6 1.5 ⇒ x = 3 × 3.6 ⇒ x = 10.8. \Rightarrow \dfrac{1.5}{4.5} = \dfrac{3.6}{x} \\[1em] \Rightarrow x = \dfrac{4.5 \times 3.6}{1.5} \\[1em] \Rightarrow x = 3 \times 3.6 \\[1em] \Rightarrow x = 10.8. ⇒ 4.5 1.5 = x 3.6 ⇒ x = 1.5 4.5 × 3.6 ⇒ x = 3 × 3.6 ⇒ x = 10.8.
Hence, the fourth proportional is 10.8.
(iv) Given,
a2 , ab and b2
Let the fourth proportional to a2 , ab and b2 be x,
⇒ a2 : ab = b2 : x
⇒ a 2 a b = b 2 x ⇒ x = a b × b 2 a 2 ⇒ x = b 3 a . \Rightarrow \dfrac{a^2}{ab} = \dfrac{b^2}{x} \\[1em] \Rightarrow x = \dfrac{ab \times b^2}{a^2} \\[1em] \Rightarrow x = \dfrac{b^3}{a}. ⇒ ab a 2 = x b 2 ⇒ x = a 2 ab × b 2 ⇒ x = a b 3 .
Hence, the fourth proportional is b 3 a \dfrac{b^3}{a} a b 3
(v) Given,
(a2 − ab + b2 ), (a3 + b3 ) and (a − b)
Let the fourth proportional to (a2 − ab + b2 ), (a3 + b3 ) and (a − b) be x,
⇒ a2 : ab = b2 :x
⇒ a 2 − a b + b 2 a 3 + b 3 = a − b x ⇒ x = ( a 3 + b 3 ) ( a − b ) a 2 − a b + b 2 ⇒ x = ( a + b ) ( a 2 − a b + b 2 ) ( a − b ) a 2 − a b + b 2 ⇒ x = ( a + b ) ( a − b ) ⇒ x = a 2 − b 2 . \Rightarrow \dfrac{a^2 - ab + b^2}{a^3 + b^3} = \dfrac{a - b}{x} \\[1em] \Rightarrow x = \dfrac{(a^3 + b^3)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = \dfrac{(a + b)(a^2 - ab + b^2)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = (a + b)(a - b) \\[1em] \Rightarrow x = a^2 - b^2. ⇒ a 3 + b 3 a 2 − ab + b 2 = x a − b ⇒ x = a 2 − ab + b 2 ( a 3 + b 3 ) ( a − b ) ⇒ x = a 2 − ab + b 2 ( a + b ) ( a 2 − ab + b 2 ) ( a − b ) ⇒ x = ( a + b ) ( a − b ) ⇒ x = a 2 − b 2 .
Hence, the fourth proportional is a2 - b2 .
Find the third proportional to :
(i) 9 and 6
(ii) 2 2 3 2\dfrac{2}{3} 2 3 2
(iii) 1.6 and 2.4
(iv) (2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
(v) ( a b + b a ) \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) ( b a + a b ) a 2 + b 2 \sqrt{a^{2} + b^{2}} a 2 + b 2
Answer
(i) Given,
9 and 6
Let third proportional to 9 and 6 be x.
⇒ 9 : 6 = 6 : x
⇒ 9 6 = 6 x \dfrac{9}{6} = \dfrac{6}{x} 6 9 = x 6
⇒ x = 6 2 9 = 36 4 \dfrac{6^2}{9} = \dfrac{36}{4} 9 6 2 = 4 36
⇒ x = 4.
Hence, the third proportional is 4.
(ii) Given,
2 2 3 2\dfrac{2}{3} 2 3 2
Let third proportional to 8 3 \dfrac{8}{3} 3 8
8 3 \dfrac{8}{3} 3 8
⇒ 4 x = 8 3 4 ⇒ 4 x = 2 3 ⇒ x = 3 2 × 4 ⇒ x = 6. \Rightarrow \dfrac{4}{x} = \dfrac{\dfrac{8}{3}}{4} \\[1em] \Rightarrow \dfrac{4}{x} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow x = 6. ⇒ x 4 = 4 3 8 ⇒ x 4 = 3 2 ⇒ x = 2 3 × 4 ⇒ x = 6.
Hence, the third proportional is 6.
(iii) Given,
1.6 and 2.4
Let third proportional to 1.6 and 2.4 be x.
1.6 : 2.4 = 2.4 : x
⇒ 1.6 2.4 = 2.4 x \dfrac{1.6}{2.4} = \dfrac{2.4}{x} 2.4 1.6 = x 2.4
⇒ x = ( 2.4 ) 2 1.6 = 5.76 1.6 \dfrac{(2.4)^2}{1.6} = \dfrac{5.76}{1.6} 1.6 ( 2.4 ) 2 = 1.6 5.76
⇒ x = 3.6
Hence, the third proportional is 3.6.
(iv) Given,
(2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
Let third proportional to (2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
( 2 + 3 ) : ( 5 + 4 3 ) = ( 5 + 4 3 ) : x (2 + \sqrt{3}) : (5 + 4 \sqrt{3}) = (5 + 4\sqrt{3}) : x ( 2 + 3 ) : ( 5 + 4 3 ) = ( 5 + 4 3 ) : x
Thus,
⇒ ( 2 + 3 ) ( 5 + 4 3 ) = ( 5 + 4 3 ) x ⇒ x = ( 5 + 4 3 ) 2 ( 2 + 3 ) = 5 2 + 2 ( 5 ) ( 4 3 ) + ( 4 3 ) 2 ( 2 + 3 ) = 25 + 40 3 + 16 × 3 ( 2 + 3 ) = 25 + 40 3 + 48 ( 2 + 3 ) = 73 + 40 3 ( 2 + 3 ) \Rightarrow \dfrac{(2 + \sqrt{3})}{(5 + 4 \sqrt{3})} = \dfrac{(5 + 4 \sqrt{3})}{x} \\[1em] \Rightarrow x = \dfrac{(5 + 4 \sqrt{3})^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{5^2 + 2(5)(4\sqrt3) + (4\sqrt3)^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 16 \times 3}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 48}{(2 + \sqrt{3})} \\[1em] = \dfrac{73 + 40\sqrt3}{(2 + \sqrt{3})} ⇒ ( 5 + 4 3 ) ( 2 + 3 ) = x ( 5 + 4 3 ) ⇒ x = ( 2 + 3 ) ( 5 + 4 3 ) 2 = ( 2 + 3 ) 5 2 + 2 ( 5 ) ( 4 3 ) + ( 4 3 ) 2 = ( 2 + 3 ) 25 + 40 3 + 16 × 3 = ( 2 + 3 ) 25 + 40 3 + 48 = ( 2 + 3 ) 73 + 40 3
Multiplying numerator and denominator by ( 2 − 3 ) (2 - \sqrt{3}) ( 2 − 3 )
= ( 73 + 40 3 ) ( 2 − 3 ) ( 2 + 3 ) ( 2 − 3 ) = 146 − 73 3 + 80 3 − 40 ( 3 ) 2 2 2 − ( 3 ) 2 = 146 + 7 3 − 120 4 − 3 = 26 + 7 3 . = \dfrac{(73 + 40\sqrt3)(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \\[1em] = \dfrac{146 - 73\sqrt3 + 80\sqrt{3} - 40(\sqrt3)^2}{2^2 - (\sqrt{3})^2} \\[1em] = \dfrac{146 + 7\sqrt3 - 120}{4 - 3} \\[1em] = 26 + 7\sqrt3. = ( 2 + 3 ) ( 2 − 3 ) ( 73 + 40 3 ) ( 2 − 3 ) = 2 2 − ( 3 ) 2 146 − 73 3 + 80 3 − 40 ( 3 ) 2 = 4 − 3 146 + 7 3 − 120 = 26 + 7 3 .
Hence, the third proportional is 26 + 7 3 26 + 7\sqrt3 26 + 7 3
(v) Given,
( a b + b a ) \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) ( b a + a b ) a 2 + b 2 \sqrt{a^{2} + b^{2}} a 2 + b 2
Let third proportional to ( a b + b a ) and a 2 + b 2 \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) \text{ and } \sqrt{a^{2} + b^{2}} ( b a + a b ) and a 2 + b 2
( a b + b a ) : a 2 + b 2 = a 2 + b 2 : x \Big(\dfrac{a}{b} + \dfrac{b}{a}\Big): \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + b^{2}}:x ( b a + a b ) : a 2 + b 2 = a 2 + b 2 : x
( a b + b a ) a 2 + b 2 = a 2 + b 2 x x = ( a 2 + b 2 ) 2 ( a b + b a ) = a 2 + b 2 a 2 + b 2 a b = ( a 2 + b 2 ) × a b a 2 + b 2 = a b . \dfrac{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)}{\sqrt{a^{2} + b^{2}}} = \dfrac{\sqrt{a^{2} + b^{2}}}{x} \\[1em] x = \dfrac{(\sqrt{a^2 + b^2})^2}{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)} \\[1em] = \dfrac{a^2 + b^2}{\dfrac{a^2 + b^2}{ab}} \\[1em] = (a^2 + b^2) \times \dfrac{ab}{a^2 + b^2} \\[1em] = ab. a 2 + b 2 ( b a + a b ) = x a 2 + b 2 x = ( b a + a b ) ( a 2 + b 2 ) 2 = ab a 2 + b 2 a 2 + b 2 = ( a 2 + b 2 ) × a 2 + b 2 ab = ab .
Hence, the third proportional is ab.
Find the mean proportion between :
(i) 28 and 63
(ii) 2.5 and 0.9
(iii) 6.25 and 1.6
(iv) ( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
(v) 6 + 3 3 \sqrt{3} 3 3 \sqrt{3} 3
Answer
(i) Given,
28 and 63
Let mean proportional between 28 and 63 be x.
28 : x :: x : 63
⇒ 28 x = x 63 ⇒ x 2 = 28 × 63 ⇒ x 2 = 1764 ⇒ x = 1764 = 42 \Rightarrow \dfrac{28}{x} = \dfrac{x}{63} \\[1em] \Rightarrow x^2 = 28 \times 63 \\[1em] \Rightarrow x^2 = 1764 \\[1em] \Rightarrow x = \sqrt{1764} = 42 ⇒ x 28 = 63 x ⇒ x 2 = 28 × 63 ⇒ x 2 = 1764 ⇒ x = 1764 = 42
Hence, the mean proportional is 42.
(ii) Given,
2.5 and 0.9
Let mean proportional between 2.5 and 0.9 be x.
2.5 : x :: x : 0.9
⇒ 2.5 x = x 0.9 ⇒ x 2 = 2.5 × 0.9 ⇒ x 2 = 2.25 ⇒ x = 2.25 = 1.5 \Rightarrow \dfrac{2.5}{x} = \dfrac{x}{0.9} \\[1em] \Rightarrow x^2 = 2.5 \times 0.9 \\[1em] \Rightarrow x^2 = 2.25 \\[1em] \Rightarrow x = \sqrt{2.25} = 1.5 ⇒ x 2.5 = 0.9 x ⇒ x 2 = 2.5 × 0.9 ⇒ x 2 = 2.25 ⇒ x = 2.25 = 1.5
Hence, the mean proportional is 1.5.
(iii) Given,
6.25 and 1.6
Let mean proportional between 6.25 and 1.6 be x.
6.25 : x :: x : 1.6
⇒ 6.25 x = x 1.6 ⇒ x 2 = 6.25 × 1.6 ⇒ x 2 = 10 ⇒ x = 10 \Rightarrow \dfrac{6.25}{x} = \dfrac{x}{1.6} \\[1em] \Rightarrow x^2 = 6.25 \times 1.6 \\[1em] \Rightarrow x^2 = 10 \\[1em] \Rightarrow x = \sqrt{10} ⇒ x 6.25 = 1.6 x ⇒ x 2 = 6.25 × 1.6 ⇒ x 2 = 10 ⇒ x = 10
Hence, the mean proportional is 10 \sqrt{10} 10
(iv) Given,
( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
Let mean proportional between ( 26 − 17 ) \Big(\sqrt{26} - \sqrt{17}\Big) ( 26 − 17 ) ( 26 + 17 ) \Big(\sqrt{26} + \sqrt{17}\Big) ( 26 + 17 )
( 26 − 17 ) : x : : x : ( 26 + 17 ) \Big(\sqrt{26} - \sqrt{17}\Big):x::x:\Big(\sqrt{26} + \sqrt{17}\Big) ( 26 − 17 ) : x :: x : ( 26 + 17 )
⇒ 26 − 17 x = x 26 + 17 ⇒ x 2 = ( 26 − 17 ) × ( 26 + 17 ) \Rightarrow \dfrac{\sqrt{26} - \sqrt{17}}{x} = \dfrac{x}{\sqrt{26} + \sqrt{17}} \\[1em] \Rightarrow x^2 = (\sqrt{26} - \sqrt{17}) \times (\sqrt{26} + \sqrt{17}) \\[1em] ⇒ x 26 − 17 = 26 + 17 x ⇒ x 2 = ( 26 − 17 ) × ( 26 + 17 )
Hence, the mean proportional is 3.
(v) Given,
6 + 3 3 3\sqrt{3} 3 3 4 3 4\sqrt{3} 4 3
Let mean proportion between 6 + 3 3 3\sqrt{3} 3 3 4 3 4\sqrt{3} 4 3
6 + 3 3 : x : : x : 8 − 4 3 \sqrt{3} : x :: x : 8 − 4 \sqrt{3} 3 : x :: x : 8 − 4 3
⇒ 6 + 3 3 x = x 8 − 4 3 ⇒ x 2 = ( 6 + 3 3 ) × ( 8 − 4 3 ) ⇒ x 2 = 48 − 24 3 + 24 3 − 36 ⇒ x 2 = 12 ⇒ x = 2 3 \Rightarrow \dfrac{6 + 3\sqrt{3}}{x} = \dfrac{x}{8 − 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3}) \times (8 − 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt3 ⇒ x 6 + 3 3 = 8 − 4 3 x ⇒ x 2 = ( 6 + 3 3 ) × ( 8 − 4 3 ) ⇒ x 2 = 48 − 24 3 + 24 3 − 36 ⇒ x 2 = 12 ⇒ x = 2 3
Hence, the mean proportional is 2 3 2\sqrt3 2 3
6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.
Answer
Let two numbers be x and y.
Given,
6 is mean proportion between x and y,
∴ x 6 = 6 y ⇒ x y = 36 . . . . . ( 1 ) \therefore \dfrac{x}{6} = \dfrac{6}{y} \\[1em] \Rightarrow xy = 36 \space .....(1) ∴ 6 x = y 6 ⇒ x y = 36 ..... ( 1 )
Given,
48 is third proportional to x and y,
∴ x y = y 48 ⇒ y 2 = 48 x ⇒ x = y 2 48 . . . . . ( 2 ) \therefore \dfrac{x}{y} = \dfrac{y}{48} \\[1em] \Rightarrow y^2 = 48x \\[1em] \Rightarrow x = \dfrac{y^2}{48} \space .....(2) ∴ y x = 48 y ⇒ y 2 = 48 x ⇒ x = 48 y 2 ..... ( 2 )
Substituting value of x from equation (2) in (1) we get,
⇒ y 2 48 . y = 36 ⇒ y 3 = 36 × 48 ⇒ y 3 = 1728 ⇒ y = 1728 3 ⇒ y = 12. \Rightarrow \dfrac{y^2}{48}.y = 36 \\[1em] \Rightarrow y^3 = 36 \times 48 \\[1em] \Rightarrow y^3 = 1728 \\[1em] \Rightarrow y = \sqrt[3]{1728} \\[1em] \Rightarrow y = 12. ⇒ 48 y 2 . y = 36 ⇒ y 3 = 36 × 48 ⇒ y 3 = 1728 ⇒ y = 3 1728 ⇒ y = 12.
Substituting value of y in equation (2), we get :
x = 12 2 48 = 144 48 x = \dfrac{12^2}{48} = \dfrac{144}{48} x = 48 1 2 2 = 48 144
Hence, numbers are 3 and 12.
What least number must be added to each of the numbers 5, 11, 19 and 37, so that the resulting numbers are proportional.
Answer
Let least number to be added to numbers be x.
∴ 5 + x : 11 + x :: 19 + x : 37 + x
⇒ 5 + x 11 + x = 19 + x 37 + x ⇒ ( 5 + x ) ( 37 + x ) = ( 19 + x ) ( 11 + x ) ⇒ 185 + 5 x + 37 x + x 2 = 209 + 19 x + 11 x + x 2 ⇒ x 2 + 42 x + 185 = x 2 + 30 x + 209 ⇒ x 2 − x 2 + 42 x − 30 x = 209 − 185 ⇒ 12 x = 24 ⇒ x = 2. \Rightarrow \dfrac{5 + x}{11 + x} = \dfrac{19 + x}{37 + x} \\[1em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[1em] \Rightarrow 185 + 5x + 37x + x^2 = 209 + 19x + 11x + x^2 \\[1em] \Rightarrow x^2 + 42x + 185 = x^2 + 30x + 209 \\[1em] \Rightarrow x^2 - x^2 + 42x - 30x = 209 - 185 \\[1em] \Rightarrow 12x = 24 \\[1em] \Rightarrow x = 2. ⇒ 11 + x 5 + x = 37 + x 19 + x ⇒ ( 5 + x ) ( 37 + x ) = ( 19 + x ) ( 11 + x ) ⇒ 185 + 5 x + 37 x + x 2 = 209 + 19 x + 11 x + x 2 ⇒ x 2 + 42 x + 185 = x 2 + 30 x + 209 ⇒ x 2 − x 2 + 42 x − 30 x = 209 − 185 ⇒ 12 x = 24 ⇒ x = 2.
Hence, least number to be added to make numbers proportional is 2.
What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion ?
Answer
Let the number to be added to numbers be x.
∴ 4 + x : 6 + x :: 8 + x : 11 + x
⇒ 4 + x 6 + x = 8 + x 11 + x ⇒ ( 4 + x ) ( 11 + x ) = ( 8 + x ) ( 6 + x ) ⇒ 44 + 4 x + 11 x + x 2 = 48 + 8 x + 6 x + x 2 ⇒ x 2 + 15 x + 44 = x 2 + 14 x + 48 ⇒ x 2 − x 2 + 15 x − 14 x = 48 − 44 ⇒ x = 4. \Rightarrow \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x)(11 + x) = (8 + x)(6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2 \\[1em] \Rightarrow x^2 + 15x + 44 = x^2 + 14x + 48 \\[1em] \Rightarrow x^2 - x^2 + 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4. ⇒ 6 + x 4 + x = 11 + x 8 + x ⇒ ( 4 + x ) ( 11 + x ) = ( 8 + x ) ( 6 + x ) ⇒ 44 + 4 x + 11 x + x 2 = 48 + 8 x + 6 x + x 2 ⇒ x 2 + 15 x + 44 = x 2 + 14 x + 48 ⇒ x 2 − x 2 + 15 x − 14 x = 48 − 44 ⇒ x = 4.
Hence, the number to be added to make numbers proportional is 4.
What least number must be subtracted from each of the numbers 23, 30, 57 and 78, so that the remainders are in proportion?
Answer
Let least number to be subtracted to numbers be x.
∴ 23 - x : 30 - x :: 57 - x : 78 - x
⇒ 23 − x 30 − x = 57 − x 78 − x ⇒ ( 23 − x ) ( 78 − x ) = ( 57 − x ) ( 30 − x ) ⇒ 1794 − 23 x − 78 x + x 2 = 1710 − 57 x − 30 x + x 2 ⇒ x 2 − 101 x + 1794 = x 2 − 87 x + 1710 ⇒ x 2 − x 2 − 101 x + 87 x = 1710 − 1794 ⇒ − 14 x = − 84 ⇒ x = − 84 − 14 ⇒ x = 6. \Rightarrow \dfrac{23 - x}{30 - x} = \dfrac{57 - x}{78 - x} \\[1em] \Rightarrow (23 - x)(78 - x) = (57 - x)(30 - x) \\[1em] \Rightarrow 1794 - 23x - 78x + x^2 = 1710 - 57x - 30x + x^2 \\[1em] \Rightarrow x^2 - 101x + 1794 = x^2 - 87x + 1710 \\[1em] \Rightarrow x^2 - x^2 - 101x + 87x = 1710 - 1794 \\[1em] \Rightarrow -14x = -84 \\[1em] \Rightarrow x = \dfrac{-84}{-14} \\[1em] \Rightarrow x = 6. ⇒ 30 − x 23 − x = 78 − x 57 − x ⇒ ( 23 − x ) ( 78 − x ) = ( 57 − x ) ( 30 − x ) ⇒ 1794 − 23 x − 78 x + x 2 = 1710 − 57 x − 30 x + x 2 ⇒ x 2 − 101 x + 1794 = x 2 − 87 x + 1710 ⇒ x 2 − x 2 − 101 x + 87 x = 1710 − 1794 ⇒ − 14 x = − 84 ⇒ x = − 14 − 84 ⇒ x = 6.
Hence, least number to be subtracted to make numbers proportional is 6.
If (x − 2), (x + 2), (2x + 1) and (2x + 19) are in proportion, find the value of x.
Answer
Let,
∴ x - 2 : x + 2 :: 2x + 1 : 2x + 19
⇒ x − 2 x + 2 = 2 x + 1 2 x + 19 ⇒ ( x − 2 ) ( 2 x + 19 ) = ( 2 x + 1 ) ( x + 2 ) ⇒ 2 x 2 + 19 x − 4 x − 38 = 2 x 2 + 4 x + x + 2 ⇒ 2 x 2 + 15 x − 38 = 2 x 2 + 5 x + 2 ⇒ 2 x 2 − 2 x 2 + 15 x − 5 x = 2 + 38 ⇒ 10 x = 40 ⇒ x = 4. \Rightarrow \dfrac{x - 2}{x + 2} = \dfrac{2x + 1}{2x + 19} \\[1em] \Rightarrow (x - 2)(2x + 19) = (2x + 1)(x + 2) \\[1em] \Rightarrow 2x^2 + 19x - 4x - 38 = 2x^2 + 4x + x + 2 \\[1em] \Rightarrow 2x^2 + 15x - 38 = 2x^2 + 5x + 2 \\[1em] \Rightarrow 2x^2 - 2x^2 + 15x - 5x = 2 + 38 \\[1em] \Rightarrow 10x = 40 \\[1em] \Rightarrow x = 4. ⇒ x + 2 x − 2 = 2 x + 19 2 x + 1 ⇒ ( x − 2 ) ( 2 x + 19 ) = ( 2 x + 1 ) ( x + 2 ) ⇒ 2 x 2 + 19 x − 4 x − 38 = 2 x 2 + 4 x + x + 2 ⇒ 2 x 2 + 15 x − 38 = 2 x 2 + 5 x + 2 ⇒ 2 x 2 − 2 x 2 + 15 x − 5 x = 2 + 38 ⇒ 10 x = 40 ⇒ x = 4.
Hence, the value of x = 4.
The following numbers, K + 3, K + 2, 3K − 7 and 2K − 3 are in proportion. Find the value of K.
Answer
Let,
∴ K + 3 : K + 2 :: 3K − 7 : 2K − 3
⇒ K + 3 K + 2 = 3 K − 7 2 K − 3 ⇒ ( K + 3 ) ( 2 K − 3 ) = ( 3 K − 7 ) ( K + 2 ) ⇒ 2 K 2 − 3 K + 6 k − 9 = 3 K 2 + 6 K − 7 K − 14 ⇒ 2 K 2 + 3 K − 9 = 3 K 2 − K − 14 ⇒ 0 = 3 K 2 − K − 14 − 2 K 2 − 3 K + 9 ⇒ K 2 − 4 K − 5 = 0 ⇒ K 2 + 1 K − 5 K − 5 = 0 ⇒ K ( K + 1 ) − 5 ( K + 1 ) = 0 ⇒ ( K − 5 ) ( K + 1 ) = 0 ⇒ ( K − 5 ) = 0 or ( K + 1 ) = 0 [Using Zero - product rule] ⇒ K = 5 or K = − 1 \Rightarrow \dfrac{K + 3}{K + 2} = \dfrac{3K - 7}{2K - 3} \\[1em] \Rightarrow (K + 3)(2K - 3) = (3K - 7)(K + 2) \\[1em] \Rightarrow 2K^2 - 3K + 6k - 9 = 3K^2 + 6K - 7K - 14 \\[1em] \Rightarrow 2K^2 + 3K - 9 = 3K^2 - K - 14 \\[1em] \Rightarrow 0 = 3K^2 - K - 14 - 2K^2 - 3K + 9 \\[1em] \Rightarrow K^2 - 4K - 5 = 0 \\[1em] \Rightarrow K^2 + 1K - 5K - 5 = 0 \\[1em] \Rightarrow K(K + 1) - 5(K + 1) = 0 \\[1em] \Rightarrow (K - 5)(K + 1) = 0 \\[1em] \Rightarrow (K - 5) = 0 \text{ or }(K + 1) = 0 \text{[Using Zero - product rule]}\\[1em] \Rightarrow K = 5 \text{ or } K = - 1 ⇒ K + 2 K + 3 = 2 K − 3 3 K − 7 ⇒ ( K + 3 ) ( 2 K − 3 ) = ( 3 K − 7 ) ( K + 2 ) ⇒ 2 K 2 − 3 K + 6 k − 9 = 3 K 2 + 6 K − 7 K − 14 ⇒ 2 K 2 + 3 K − 9 = 3 K 2 − K − 14 ⇒ 0 = 3 K 2 − K − 14 − 2 K 2 − 3 K + 9 ⇒ K 2 − 4 K − 5 = 0 ⇒ K 2 + 1 K − 5 K − 5 = 0 ⇒ K ( K + 1 ) − 5 ( K + 1 ) = 0 ⇒ ( K − 5 ) ( K + 1 ) = 0 ⇒ ( K − 5 ) = 0 or ( K + 1 ) = 0 [Using Zero - product rule] ⇒ K = 5 or K = − 1
Hence, K = 5 or K = −1.
If (x + 5) is the geometric mean between (x + 2) and (x + 9), find the value of x.
Answer
Given, x + 5 is G.M. between x + 2 and x + 9.
∴ x + 2 x + 5 = x + 5 x + 9 \therefore \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} ∴ x + 5 x + 2 = x + 9 x + 5
⇒ (x + 5)2 = (x + 2)(x + 9)
⇒ x2 + 10x + 25 = x2 + 9x + 2x + 18
⇒ x2 + 10x + 25 = x2 + 11x + 18
⇒ x2 - x2 + 10x - 11x = 18 - 25
⇒ -x = -7
⇒ x = 7.
Hence, the value of x = 7.
Find two numbers whose mean proportion is 36 and the third proportional is 288.
Answer
Let the two numbers be x and y.
Thus, 36 is the mean proportion between x and y.
⇒ x 36 = 36 y ⇒ x y = 36 2 ⇒ x y = 1296 ⇒ x = 1296 y .....(1) \Rightarrow \dfrac{x}{36} = \dfrac{36}{y} \\[1em] \Rightarrow xy = 36^2 \\[1em] \Rightarrow xy = 1296 \\[1em] \Rightarrow x = \dfrac{1296}{y}\text{.....(1)} ⇒ 36 x = y 36 ⇒ x y = 3 6 2 ⇒ x y = 1296 ⇒ x = y 1296 .....(1)
The third proportional for x and y is 288.
⇒ x y = y 288 ⇒ y 2 = 288 x ..........(2) \Rightarrow \dfrac{x}{y} = \dfrac{y}{288} \\[1em] \Rightarrow y^2 = 288x \text{ ..........(2)} ⇒ y x = 288 y ⇒ y 2 = 288 x ..........(2)
Substituting value of x from equation (1) in (2), we get :
⇒ y 2 = 288 × 1296 y ⇒ y 3 = 373248 ⇒ y = 373248 3 = 72. \Rightarrow y^2 = 288 \times \dfrac{1296}{y} \\[1em] \Rightarrow y^3 = 373248 \\[1em] \Rightarrow y = \sqrt[3]{373248} = 72. ⇒ y 2 = 288 × y 1296 ⇒ y 3 = 373248 ⇒ y = 3 373248 = 72.
Substituting the value of y in equation (1), we get :
⇒ x = 1296 y ⇒ x = 1296 72 ⇒ x = 18 \Rightarrow x = \dfrac{1296}{y} \\[1em] \Rightarrow x = \dfrac{1296}{72} \\[1em] \Rightarrow x = 18 ⇒ x = y 1296 ⇒ x = 72 1296 ⇒ x = 18
Hence, the two numbers are 18 and 72.
If a : b :: c : d, prove that :
(i) (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd)
(ii) (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
(iii) (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 )
Answer
(i) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd), we get :
⇒ a 2 + a b c 2 + c d ⇒ ( c k ) 2 + c k . d k c 2 + c d ⇒ k 2 c 2 + k 2 c d c 2 + c d ⇒ k 2 ( c 2 + c d ) c 2 + c d ⇒ k 2 . \Rightarrow \dfrac{a^2 + ab}{c^2 + cd} \\[1em] \Rightarrow \dfrac{(ck)^2 + ck.dk}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd)}{c^2 + cd} \\[1em] \Rightarrow k^2. ⇒ c 2 + c d a 2 + ab ⇒ c 2 + c d ( c k ) 2 + c k . d k ⇒ c 2 + c d k 2 c 2 + k 2 c d ⇒ c 2 + c d k 2 ( c 2 + c d ) ⇒ k 2 .
Substituting value of a and b in R.H.S. of (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd), we get :
⇒ b 2 − 2 a b d 2 − 2 c d ⇒ ( k d ) 2 − 2 × c k × d k d 2 − 2 c d ⇒ k 2 d 2 − 2 k 2 c d d 2 − 2 c d ⇒ k 2 ( d 2 − 2 c d ) d 2 − 2 c d ⇒ k 2 . \Rightarrow \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{(kd)^2 - 2 \times ck \times dk}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2 d^2 - 2k^2 cd}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2(d^2 - 2cd)}{d^2 - 2cd} \\[1em] \Rightarrow k^2. ⇒ d 2 − 2 c d b 2 − 2 ab ⇒ d 2 − 2 c d ( k d ) 2 − 2 × c k × d k ⇒ d 2 − 2 c d k 2 d 2 − 2 k 2 c d ⇒ d 2 − 2 c d k 2 ( d 2 − 2 c d ) ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd).
(ii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
⇒ a 2 + b 2 c 2 + d 2 ⇒ ( k c ) 2 + ( k d ) 2 c 2 + d 2 ⇒ k 2 c 2 + k 2 d 2 c 2 + d 2 ⇒ k 2 ( c 2 + d 2 ) c 2 + d 2 ⇒ k 2 . \Rightarrow \dfrac{a^2 + b^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kd)^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{k^2c^2 + k^2d^2}{c^2 + d^2} \\[1em] \Rightarrow \dfrac{k^2(c^2 + d^2)}{c^2 + d^2} \\[1em] \Rightarrow k^2. ⇒ c 2 + d 2 a 2 + b 2 ⇒ c 2 + d 2 ( k c ) 2 + ( k d ) 2 ⇒ c 2 + d 2 k 2 c 2 + k 2 d 2 ⇒ c 2 + d 2 k 2 ( c 2 + d 2 ) ⇒ k 2 .
Substituting value of a and b in R.H.S. of (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc)
⇒ a b + a d − b c c d − a d + b c ⇒ ( k c ) ( k d ) + ( k c ) d − ( k d ) c c d − ( k c ) d + ( k d ) c ⇒ k 2 c d + k c d − k c d c d − k c d + k c d ⇒ k 2 ( c d ) ( c d ) ⇒ k 2 . \Rightarrow \dfrac{ab + ad - bc}{cd - ad + bc} \\[1em] \Rightarrow \dfrac{(kc)(kd) + (kc)d - (kd)c}{cd - (kc)d + (kd)c} \\[1em] \Rightarrow \dfrac{k^2 cd + kcd - kcd}{cd - kcd + kcd} \\[1em] \Rightarrow \dfrac{k^2 (cd)}{(cd)} \\[1em] \Rightarrow k^2. ⇒ c d − a d + b c ab + a d − b c ⇒ c d − ( k c ) d + ( k d ) c ( k c ) ( k d ) + ( k c ) d − ( k d ) c ⇒ c d − k c d + k c d k 2 c d + k c d − k c d ⇒ ( c d ) k 2 ( c d ) ⇒ k 2 .
Since. L.H.S. = R.H.S.
Hence, proved that (a2 + b2 ) : (c2 + d2 ) = (ab + ad − bc) : (cd − ad + bc).
(iii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ),
⇒ a 2 + a c + c 2 a 2 − a c + c 2 ⇒ ( k c ) 2 + ( k c ) c + c 2 ( k c ) 2 − ( k c ) c + c 2 ⇒ k 2 c 2 + k c 2 + c 2 k 2 c 2 − k c 2 + c 2 ⇒ c 2 ( k 2 + k + 1 ) c 2 ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{a^2 + ac + c^2}{a^2 - ac + c^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)c + c^2}{(kc)^2 - (kc)c + c^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + kc^2 + c^2}{k^2 c^2 - kc^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)}{c^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a 2 − a c + c 2 a 2 + a c + c 2 ⇒ ( k c ) 2 − ( k c ) c + c 2 ( k c ) 2 + ( k c ) c + c 2 ⇒ k 2 c 2 − k c 2 + c 2 k 2 c 2 + k c 2 + c 2 ⇒ c 2 ( k 2 − k + 1 ) c 2 ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Substituting value of a and b in R.H.S. of (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ),
⇒ b 2 + b d + d 2 b 2 − b d + d 2 ⇒ ( k d ) 2 + ( k d ) d + d 2 ( k d ) 2 − ( k d ) d + d 2 ⇒ k 2 d 2 + k d 2 + d 2 k 2 d 2 − k d 2 + d 2 ⇒ d 2 ( k 2 + k + 1 ) d 2 ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{b^2 + bd + d^2}{b^2 - bd + d^2} \\[1em] \Rightarrow \dfrac{(kd)^2 + (kd)d + d^2}{(kd)^2 - (kd)d + d^2} \\[1em] \Rightarrow \dfrac{k^2 d^2 + kd^2 + d^2}{k^2 d^2 - kd^2 + d^2} \\[1em] \Rightarrow \dfrac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ b 2 − b d + d 2 b 2 + b d + d 2 ⇒ ( k d ) 2 − ( k d ) d + d 2 ( k d ) 2 + ( k d ) d + d 2 ⇒ k 2 d 2 − k d 2 + d 2 k 2 d 2 + k d 2 + d 2 ⇒ d 2 ( k 2 − k + 1 ) d 2 ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 + ac + c2 ) : (a2 − ac + c2 ) = (b2 + bd + d2 ) : (b2 − bd + d2 ).
If a : b :: c : d, show that :
(i) a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
(ii) m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
(iii) a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
(iv) ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
Answer
(i) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
⇒ a + b c + d ⇒ k c + k d c + d ⇒ k ( c + d ) ( c + d ) ⇒ k . \Rightarrow \dfrac{a + b}{c + d} \\[1em] \Rightarrow \dfrac{kc + kd}{c + d} \\[1em] \Rightarrow \dfrac{k(c + d)}{(c + d)} \\[1em] \Rightarrow k. ⇒ c + d a + b ⇒ c + d k c + k d ⇒ ( c + d ) k ( c + d ) ⇒ k .
Substituting value of a and b in R.H.S. of a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
⇒ 2 a 2 + 7 b 2 2 c 2 + 7 d 2 ⇒ 2 ( k c ) 2 + 7 ( d k ) 2 2 c 2 + 7 d 2 ⇒ k 2 ( 2 c 2 + 7 d 2 ) 2 c 2 + 7 d 2 ⇒ k 2 ⇒ k . \Rightarrow \sqrt{\dfrac{2a^2 + 7b^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{2(kc)^2 + 7(dk)^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{k^2(2c^2 + 7d^2)}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{k^2} \\[1em] \Rightarrow k. ⇒ 2 c 2 + 7 d 2 2 a 2 + 7 b 2 ⇒ 2 c 2 + 7 d 2 2 ( k c ) 2 + 7 ( d k ) 2 ⇒ 2 c 2 + 7 d 2 k 2 ( 2 c 2 + 7 d 2 ) ⇒ k 2 ⇒ k .
Since, L.H.S. = R.H.S.
Hence, proved that a + b c + d = 2 a 2 + 7 b 2 2 c 2 + 7 d 2 \dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}} c + d a + b = 2 c 2 + 7 d 2 2 a 2 + 7 b 2
(ii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting value of a and b in L.H.S. of m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
⇒ m a 2 + n c 2 m b 2 + n d 2 ⇒ m ( k c ) 2 + n c 2 m ( k d ) 2 + n d 2 ⇒ m k 2 c 2 + n c 2 m k 2 d 2 + n d 2 ⇒ c 2 ( m k 2 + n ) d 2 ( m k 2 + n ) ⇒ c 2 d 2 . \Rightarrow \dfrac{ma^2 + nc^2}{mb^2 + nd^2} \\[1em] \Rightarrow \dfrac{m(kc)^2 + nc^2}{m(kd)^2 + nd^2} \\[1em] \Rightarrow \dfrac{m k^2 c^2 + n c^2}{m k^2 d^2 + n d^2} \\[1em] \Rightarrow \dfrac{c^2(m k^2 + n)}{d^2(m k^2 + n)} \\[1em] \Rightarrow \dfrac{c^2}{d^2}. ⇒ m b 2 + n d 2 m a 2 + n c 2 ⇒ m ( k d ) 2 + n d 2 m ( k c ) 2 + n c 2 ⇒ m k 2 d 2 + n d 2 m k 2 c 2 + n c 2 ⇒ d 2 ( m k 2 + n ) c 2 ( m k 2 + n ) ⇒ d 2 c 2 .
Substituting value of a and b in R.H.S. of m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
⇒ a 4 + c 4 b 4 + d 4 ⇒ ( k c ) 4 + c 4 ( k d ) 4 + d 4 ⇒ k 4 c 4 + c 4 k 4 d 4 + d 4 ⇒ c 4 ( k 4 + 1 ) d 4 ( k 4 + 1 ) ⇒ c 4 d 4 ⇒ c 2 d 2 . \Rightarrow \sqrt{\dfrac{a^4 + c^4}{b^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{(kc)^4 + c^4}{(kd)^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{k^4 c^4 + c^4}{k^4 d^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4(k^4 + 1)}{d^4(k^4 + 1)}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4}{d^4}} \\[1em] \Rightarrow \dfrac{c^2}{d^2}. ⇒ b 4 + d 4 a 4 + c 4 ⇒ ( k d ) 4 + d 4 ( k c ) 4 + c 4 ⇒ k 4 d 4 + d 4 k 4 c 4 + c 4 ⇒ d 4 ( k 4 + 1 ) c 4 ( k 4 + 1 ) ⇒ d 4 c 4 ⇒ d 2 c 2 .
Since, L.H.S. = R.H.S.
Hence, proved that m a 2 + n c 2 m b 2 + n d 2 = a 4 + c 4 b 4 + d 4 \dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}} m b 2 + n d 2 m a 2 + n c 2 = b 4 + d 4 a 4 + c 4
(iii) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting values of a and b in L.H.S. of a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
⇒ a 2 + a b + b 2 a 2 − a b + b 2 ⇒ ( k c ) 2 + ( k c ) ( k d ) + ( k d ) 2 ( k c ) 2 − ( k c ) ( k d ) + ( k d ) 2 ⇒ k 2 c 2 + k 2 c d + k 2 d 2 k 2 c 2 − k 2 c d + k 2 d 2 ⇒ k 2 ( c 2 + c d + d 2 ) k 2 ( c 2 − c d + d 2 ) ⇒ c 2 + c d + d 2 c 2 − c d + d 2 . \Rightarrow \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)(kd) + (kd)^2}{(kc)^2 - (kc)(kd) + (kd)^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd + k^2 d^2}{k^2 c^2 - k^2 cd + k^2 d^2} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd + d^2)}{k^2(c^2 - cd + d^2)} \\[1em] \Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}. ⇒ a 2 − ab + b 2 a 2 + ab + b 2 ⇒ ( k c ) 2 − ( k c ) ( k d ) + ( k d ) 2 ( k c ) 2 + ( k c ) ( k d ) + ( k d ) 2 ⇒ k 2 c 2 − k 2 c d + k 2 d 2 k 2 c 2 + k 2 c d + k 2 d 2 ⇒ k 2 ( c 2 − c d + d 2 ) k 2 ( c 2 + c d + d 2 ) ⇒ c 2 − c d + d 2 c 2 + c d + d 2 .
Substituting values of a and b in R.H.S. of a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
⇒ c 2 + c d + d 2 c 2 − c d + d 2 . \Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}. ⇒ c 2 − c d + d 2 c 2 + c d + d 2 .
Since, L.H.S. = R.H.S.
Hence, proved that a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 \dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}} a 2 − ab + b 2 a 2 + ab + b 2 = c 2 − c d + d 2 c 2 + c d + d 2
(iv) Given,
⇒ a : b :: c : d
∴ a : b = c : d
∴ a b = c d ⇒ a c = b d = k (let) \therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)} ∴ b a = d c ⇒ c a = d b = k (let)
⇒ a = ck and b = dk.
Substituting values of a and b in L.H.S. of ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
⇒ ( a + c ) 3 ( b + d ) 3 ⇒ ( k c + c ) 3 ( k d + d ) 3 ⇒ c 3 ( k + 1 ) 3 d 3 ( k + 1 ) 3 ⇒ c 3 d 3 . \Rightarrow \dfrac{(a + c)^3}{(b + d)^3} \\[1em] \Rightarrow \dfrac{(kc + c)^3}{(kd + d)^3} \\[1em] \Rightarrow \dfrac{c^3(k + 1)^3}{d^3(k + 1)^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}. ⇒ ( b + d ) 3 ( a + c ) 3 ⇒ ( k d + d ) 3 ( k c + c ) 3 ⇒ d 3 ( k + 1 ) 3 c 3 ( k + 1 ) 3 ⇒ d 3 c 3 .
Substituting values of a and b in R.H.S. of ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
⇒ a ( a − c ) 2 b ( b − d ) 2 ⇒ k c ( k c − c ) 2 k d ( k d − d ) 2 ⇒ k c ⋅ c 2 ( k − 1 ) 2 k d ⋅ d 2 ( k − 1 ) 2 ⇒ k c 3 k d 3 ⇒ c 3 d 3 . \Rightarrow \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \Rightarrow \dfrac{kc(kc - c)^2}{kd(kd - d)^2} \\[1em] \Rightarrow \dfrac{kc \cdot c^2(k - 1)^2}{kd \cdot d^2(k - 1)^2} \\[1em] \Rightarrow \dfrac{kc^3}{kd^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}. ⇒ b ( b − d ) 2 a ( a − c ) 2 ⇒ k d ( k d − d ) 2 k c ( k c − c ) 2 ⇒ k d ⋅ d 2 ( k − 1 ) 2 k c ⋅ c 2 ( k − 1 ) 2 ⇒ k d 3 k c 3 ⇒ d 3 c 3 .
Since, L.H.S. = R.H.S.
Hence, proved that ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 \dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}} ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 2
If x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
(i) x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
(ii) x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
(iii) x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
(iv) a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) = 3 \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3 ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x = 3
Answer
(i) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
⇒ x 2 + y 2 + z 2 a 2 + b 2 + c 2 ⇒ k 2 a 2 + k 2 b 2 + k 2 c 2 a 2 + b 2 + c 2 ⇒ k 2 ( a 2 + b 2 + c 2 ) a 2 + b 2 + c 2 ⇒ k 2 . \Rightarrow \dfrac{x^2 + y^2 + z^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 a^2 + k^2 b^2 + k^2 c^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 (a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \\[1em] \Rightarrow k^2. ⇒ a 2 + b 2 + c 2 x 2 + y 2 + z 2 ⇒ a 2 + b 2 + c 2 k 2 a 2 + k 2 b 2 + k 2 c 2 ⇒ a 2 + b 2 + c 2 k 2 ( a 2 + b 2 + c 2 ) ⇒ k 2 .
Substituting values of x, y and z in R.H.S. of the equation x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
⇒ ( p x + q y + r z p a + q b + r c ) 2 ⇒ ( p ( k a ) + q ( k b ) + r ( k c ) p a + q b + r c ) 2 ⇒ ( k ( p a + q b + r c ) p a + q b + r c ) 2 ⇒ k 2 . \Rightarrow \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{p(ka) + q(kb) + r(kc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{k(pa + qb + rc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow k^2. ⇒ ( p a + q b + rc p x + q y + rz ) 2 ⇒ ( p a + q b + rc p ( ka ) + q ( kb ) + r ( k c ) ) 2 ⇒ ( p a + q b + rc k ( p a + q b + rc ) ) 2 ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that x 2 + y 2 + z 2 a 2 + b 2 + c 2 = ( p x + q y + r z p a + q b + r c ) 2 \dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2} a 2 + b 2 + c 2 x 2 + y 2 + z 2 = ( p a + q b + rc p x + q y + rz ) 2
(ii) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
⇒ x 3 a 3 + y 3 b 3 + z 3 c 3 ⇒ k 3 a 3 a 3 + k 3 b 3 b 3 + k 3 c 3 c 3 ⇒ k 3 + k 3 + k 3 ⇒ 3 k 3 . \Rightarrow \dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^3} + \dfrac{k^3 b^3}{b^3} + \dfrac{k^3 c^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3. ⇒ a 3 x 3 + b 3 y 3 + c 3 z 3 ⇒ a 3 k 3 a 3 + b 3 k 3 b 3 + c 3 k 3 c 3 ⇒ k 3 + k 3 + k 3 ⇒ 3 k 3 .
Substituting values of x, y and z in R.H.S. of the equation x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
⇒ 3 x y z a b c ⇒ 3 ( k a ) ( k b ) ( k c ) a b c ⇒ 3 k 3 a b c a b c ⇒ 3 k 3 . \Rightarrow \dfrac{3xyz}{abc} \\[1em] \Rightarrow \dfrac{3(ka)(kb)(kc)}{abc} \\[1em] \Rightarrow \dfrac{3k^3 abc}{abc} \\[1em] \Rightarrow 3k^3. ⇒ ab c 3 x yz ⇒ ab c 3 ( ka ) ( kb ) ( k c ) ⇒ ab c 3 k 3 ab c ⇒ 3 k 3 .
Since, L.H.S. = R.H.S.
Hence, proved that x 3 a 3 + y 3 b 3 + z 3 c 3 = 3 x y z a b c \dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc} a 3 x 3 + b 3 y 3 + c 3 z 3 = ab c 3 x yz
(iii) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in L.H.S. of the equation x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
⇒ x 3 a 2 + y 3 b 2 + z 3 c 2 ⇒ k 3 a 3 a 2 + k 3 b 3 b 2 + k 3 c 3 c 2 ⇒ k 3 a + k 3 b + k 3 c ⇒ k 3 ( a + b + c ) . \Rightarrow \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^2} + \dfrac{k^3 b^3}{b^2} + \dfrac{k^3 c^3}{c^2} \\[1em] \Rightarrow k^3 a + k^3 b + k^3 c \\[1em] \Rightarrow k^3 (a + b + c). ⇒ a 2 x 3 + b 2 y 3 + c 2 z 3 ⇒ a 2 k 3 a 3 + b 2 k 3 b 3 + c 2 k 3 c 3 ⇒ k 3 a + k 3 b + k 3 c ⇒ k 3 ( a + b + c ) .
Substituting values of x, y and z in R.H.S. of the equation x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
⇒ ( x + y + z ) 3 ( a + b + c ) 2 ⇒ ( k ( a + b + c ) ) 3 ( a + b + c ) 2 ⇒ k 3 ( a + b + c ) 3 ( a + b + c ) 2 ⇒ k 3 ( a + b + c ) . \Rightarrow \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{(k(a + b + c))^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{k^3 (a + b + c)^3}{(a + b + c)^2} \\[1em] \Rightarrow k^3 (a + b + c). ⇒ ( a + b + c ) 2 ( x + y + z ) 3 ⇒ ( a + b + c ) 2 ( k ( a + b + c ) ) 3 ⇒ ( a + b + c ) 2 k 3 ( a + b + c ) 3 ⇒ k 3 ( a + b + c ) .
Since, L.H.S. = R.H.S.
Hence, proved that x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 \dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}} a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3
(iv) Given,
⇒ x a = y b = z c \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} a x = b y = c z
⇒ x = ak, y = bk, z = ck.
Substituting values of x, y and z in a x − b y ( a + b ) ( x − y ) \dfrac{ax - by}{(a + b)(x - y)} ( a + b ) ( x − y ) a x − b y
⇒ a x − b y ( a + b ) ( x − y ) ⇒ a ( k a ) − b ( k b ) ( a + b ) ( k a − k b ) ⇒ k ( a 2 − b 2 ) k ( a + b ) ( a − b ) ⇒ k ( a − b ) ( a + b ) k ( a + b ) ( a − b ) ⇒ 1.... ( 1 ) \Rightarrow \dfrac{ax - by}{(a + b)(x - y)} \\[1em] \Rightarrow \dfrac{a(ka) - b(kb)}{(a + b)(ka - kb)} \\[1em] \Rightarrow \dfrac{k(a^2 - b^2)}{k(a + b)(a - b)} \\[1em] \Rightarrow \dfrac{k(a - b)(a + b)}{k(a + b)(a - b)} \\[1em] \Rightarrow 1....(1) ⇒ ( a + b ) ( x − y ) a x − b y ⇒ ( a + b ) ( ka − kb ) a ( ka ) − b ( kb ) ⇒ k ( a + b ) ( a − b ) k ( a 2 − b 2 ) ⇒ k ( a + b ) ( a − b ) k ( a − b ) ( a + b ) ⇒ 1.... ( 1 )
Substituting values of x, y and z in b y − c z ( b + c ) ( y − z ) \dfrac{by - cz}{(b + c)(y - z)} ( b + c ) ( y − z ) b y − cz
⇒ b y − c z ( b + c ) ( y − z ) ⇒ b ( k b ) − c ( k c ) ( b + c ) ( k b − k c ) ⇒ k ( b 2 − c 2 ) k ( b + c ) ( b − c ) ⇒ k ( b − c ) ( b + c ) k ( b + c ) ( b − c ) ⇒ 1.... ( 2 ) \Rightarrow \dfrac{by - cz}{(b + c)(y - z)} \\[1em] \Rightarrow \dfrac{b(kb) - c(kc)}{(b + c)(kb - kc)} \\[1em] \Rightarrow \dfrac{k(b^2 - c^2)}{k(b + c)(b - c)} \\[1em] \Rightarrow \dfrac{k(b - c)(b + c)}{k(b + c)(b - c)} \\[1em] \Rightarrow 1....(2) ⇒ ( b + c ) ( y − z ) b y − cz ⇒ ( b + c ) ( kb − k c ) b ( kb ) − c ( k c ) ⇒ k ( b + c ) ( b − c ) k ( b 2 − c 2 ) ⇒ k ( b + c ) ( b − c ) k ( b − c ) ( b + c ) ⇒ 1.... ( 2 )
Substituting values of x, y and z in c z − a x ( c + a ) ( z − x ) \dfrac{cz - ax}{(c + a)(z - x)} ( c + a ) ( z − x ) cz − a x
⇒ c z − a x ( c + a ) ( z − x ) ⇒ c ( k c ) − a ( k a ) ( c + a ) ( k c − k a ) ⇒ k ( c 2 − a 2 ) k ( c + a ) ( c − a ) ⇒ k ( c − a ) ( c + a ) k ( c + a ) ( c − a ) ⇒ 1.... ( 3 ) \Rightarrow \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow \dfrac{c(kc) - a(ka)}{(c + a)(kc - ka)} \\[1em] \Rightarrow \dfrac{k(c^2 - a^2)}{k(c + a)(c - a)} \\[1em] \Rightarrow \dfrac{k(c - a)(c + a)}{k(c + a)(c - a)} \\[1em] \Rightarrow 1....(3) ⇒ ( c + a ) ( z − x ) cz − a x ⇒ ( c + a ) ( k c − ka ) c ( k c ) − a ( ka ) ⇒ k ( c + a ) ( c − a ) k ( c 2 − a 2 ) ⇒ k ( c + a ) ( c − a ) k ( c − a ) ( c + a ) ⇒ 1.... ( 3 )
Adding 1,2 and 3 we get,
⇒ a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) ⇒ 1 + 1 + 1 ⇒ 3. \Rightarrow \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3. ⇒ ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x ⇒ 1 + 1 + 1 ⇒ 3.
Since, L.H.S = R.H.S
Hence, proved that a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) = 3 \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3 ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x = 3
If a b = c d = e f \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} b a = d c = f e
(i) (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2
(ii) a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
(iii) ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
(iv) (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
Answer
(i) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 , we get:
⇒ ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) ⇒ ( b 2 + d 2 + f 2 ) ( k 2 b 2 + k 2 d 2 + k 2 f 2 ) ⇒ ( b 2 + d 2 + f 2 ) [ k 2 ( b 2 + d 2 + f 2 ) ] ⇒ k 2 ( b 2 + d 2 + f 2 ) 2 . \Rightarrow (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)(k^2 b^2 + k^2 d^2 + k^2 f^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)[k^2(b^2 + d^2 + f^2)] \\[1em] \Rightarrow k^2(b^2 + d^2 + f^2)^2. ⇒ ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) ⇒ ( b 2 + d 2 + f 2 ) ( k 2 b 2 + k 2 d 2 + k 2 f 2 ) ⇒ ( b 2 + d 2 + f 2 ) [ k 2 ( b 2 + d 2 + f 2 )] ⇒ k 2 ( b 2 + d 2 + f 2 ) 2 .
Substituting values of a,c and e in R.H.S. of (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 , we get:
⇒ a b + c d + e f ⇒ ( k b ) b + ( k d ) d + ( k f ) f ⇒ k ( b 2 + d 2 + f 2 ) ∴ ( a b + c d + e f ) 2 = k 2 ( b 2 + d 2 + f 2 ) 2 . \Rightarrow ab + cd + ef \\[1em] \Rightarrow (k b)b + (k d)d + (k f)f \\[1em] \Rightarrow k(b^2 + d^2 + f^2) \\[1em] \therefore (ab + cd + ef)^2 = k^2(b^2 + d^2 + f^2)^2. ⇒ ab + c d + e f ⇒ ( kb ) b + ( k d ) d + ( k f ) f ⇒ k ( b 2 + d 2 + f 2 ) ∴ ( ab + c d + e f ) 2 = k 2 ( b 2 + d 2 + f 2 ) 2 .
Since, L.H.S. = R.H.S.
Hence, proved that (b2 + d2 + f2 )(a2 + c2 + e2 ) = (ab + cd + ef)2 .
(ii) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
⇒ a 3 + c 3 + e 3 b 3 + d 3 + f 3 ⇒ k 3 b 3 + k 3 d 3 + k 3 f 3 b 3 + d 3 + f 3 ⇒ k 3 ( b 3 + d 3 + f 3 ) b 3 + d 3 + f 3 ⇒ k 3 . \Rightarrow \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3 b^3 + k^3 d^3 + k^3 f^3}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3(b^3 + d^3 + f^3)}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow k^3. ⇒ b 3 + d 3 + f 3 a 3 + c 3 + e 3 ⇒ b 3 + d 3 + f 3 k 3 b 3 + k 3 d 3 + k 3 f 3 ⇒ b 3 + d 3 + f 3 k 3 ( b 3 + d 3 + f 3 ) ⇒ k 3 .
SSubstituting values of a,c and e in R.H.S. of a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
⇒ a c e b d f ⇒ ( k b ) ( k d ) ( k f ) b d f ⇒ k 3 . \Rightarrow \dfrac{ace}{bdf} \\[1em] \Rightarrow \dfrac{(k b)(k d)(k f)}{b d f} \\[1em] \Rightarrow k^3. ⇒ b df a ce ⇒ b df ( kb ) ( k d ) ( k f ) ⇒ k 3 .
Since L.H.S. = R.H.S.
Hence, proved that a 3 + c 3 + e 3 b 3 + d 3 + f 3 = a c e b d f \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf} b 3 + d 3 + f 3 a 3 + c 3 + e 3 = b df a ce
(iii) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a,c and e in L.H.S. of ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
⇒ a 2 b 2 + c 2 d 2 + e 2 f 2 ⇒ k 2 b 2 b 2 + k 2 d 2 d 2 + k 2 f 2 f 2 ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 . \Rightarrow \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \\[1em] \Rightarrow \dfrac{k^2 b^2}{b^2} + \dfrac{k^2 d^2}{d^2} + \dfrac{k^2 f^2}{f^2} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2. ⇒ b 2 a 2 + d 2 c 2 + f 2 e 2 ⇒ b 2 k 2 b 2 + d 2 k 2 d 2 + f 2 k 2 f 2 ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 .
Substituting values of a,c and e in R.H.S. of ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
⇒ a c b d + c e d f + a e b f ⇒ ( k b ) ( k d ) b d + ( k d ) ( k f ) d f + ( k b ) ( k f ) b f ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 . \Rightarrow \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\[1em] \Rightarrow \dfrac{(k b)(k d)}{b d} + \dfrac{(k d)(k f)}{d f} + \dfrac{(k b)(k f)}{b f} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2. ⇒ b d a c + df ce + b f a e ⇒ b d ( kb ) ( k d ) + df ( k d ) ( k f ) + b f ( kb ) ( k f ) ⇒ k 2 + k 2 + k 2 ⇒ 3 k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that ( a 2 b 2 + c 2 d 2 + e 2 f 2 ) = ( a c b d + c e d f + a e b f ) \Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big) ( b 2 a 2 + d 2 c 2 + f 2 e 2 ) = ( b d a c + df ce + b f a e )
(iv) Given,
⇒ a b = c d = e f = k \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k b a = d c = f e = k
⇒ a = kb, c = kd, e = kf.
Substituting values of a, c and d in L.H.S. of (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
⇒ b d f ⋅ ( a + b b + c + d d + e + f f ) 3 ⇒ b d f ⋅ ( k b + b b + k d + d d + k f + f f ) 3 ⇒ b d f ⋅ [ ( k + 1 ) + ( k + 1 ) + ( k + 1 ) ] ⇒ b d f ⋅ [ 3 ( k + 1 ) ] 3 ⇒ b d f ⋅ 27 ( k + 1 ) 3 . \Rightarrow bdf \cdot \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} \\[1em] \Rightarrow bdf \cdot \Big(\dfrac{kb + b}{b} + \dfrac{kd + d}{d} + \dfrac{kf + f}{f}\Big)^3 \\[1em] \Rightarrow bdf \cdot [(k + 1) + (k + 1) + (k + 1)] \\[1em] \Rightarrow bdf \cdot [3(k + 1)]^3 \\[1em] \Rightarrow bdf\cdot 27(k + 1)^3. ⇒ b df ⋅ ( b a + b + d c + d + f e + f ) 3 ⇒ b df ⋅ ( b kb + b + d k d + d + f k f + f ) 3 ⇒ b df ⋅ [( k + 1 ) + ( k + 1 ) + ( k + 1 )] ⇒ b df ⋅ [ 3 ( k + 1 ) ] 3 ⇒ b df ⋅ 27 ( k + 1 ) 3 .
Substituting values of a, c and d in R.H.S. of (bdf)·( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
⇒ 27 ( a + b ) ( c + d ) ( e + f ) ⇒ 27 ( k b + b ) ( k d + d ) ( k f + f ) ⇒ 27 ( b ( k + 1 ) ) ( d ( k + 1 ) ) ( f ( k + 1 ) ) ⇒ 27 b d f ( k + 1 ) 3 . \Rightarrow 27(a + b)(c + d)(e + f) \\[1em] \Rightarrow 27(kb + b)(kd + d)(kf + f) \\[1em] \Rightarrow 27\big(b(k + 1)\big)\big(d(k + 1)\big)\big(f(k + 1)\big) \\[1em] \Rightarrow 27 bdf(k + 1)^3. ⇒ 27 ( a + b ) ( c + d ) ( e + f ) ⇒ 27 ( kb + b ) ( k d + d ) ( k f + f ) ⇒ 27 ( b ( k + 1 ) ) ( d ( k + 1 ) ) ( f ( k + 1 ) ) ⇒ 27 b df ( k + 1 ) 3 .
Since, L.H.S. = R.H.S.
Hence, proved that (bdf).( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f) ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
If a, b, c are in continued proportion, prove that :
(i) a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
(ii) a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
(iii) a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
(iv) (a + b + c)(a − b + c) = (a2 + b2 + c2 )
(v) a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 )
(vi) ad(c2 + d2 ) = c3 (b + d)
Answer
(i) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
⇒ a + b b + c ⇒ c k 2 + c k c k + c ⇒ c k ( k + 1 ) c ( k + 1 ) ⇒ k . \Rightarrow \dfrac{a + b}{b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck}{ck + c} \\[1em] \Rightarrow \dfrac{c k(k + 1)}{c(k + 1)} \\[1em] \Rightarrow k. ⇒ b + c a + b ⇒ c k + c c k 2 + c k ⇒ c ( k + 1 ) c k ( k + 1 ) ⇒ k .
Substituting values of a and b in R.H.S. of equation a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
⇒ a 2 ( b − c ) b 2 ( a − b ) ⇒ ( c k 2 ) 2 ( c k − c ) ( c k ) 2 ( c k 2 − c k ) ⇒ c 2 k 4 ( c k − c ) c 2 k 2 ( c k 2 − c k ) ⇒ c 3 k 4 ( k − 1 ) c 3 k 3 ( k − 1 ) ⇒ k . \Rightarrow \dfrac{a^2(b - c)}{b^2(a - b)} \\[1em] \Rightarrow \dfrac{(ck^2)^2\big(ck - c\big)}{(ck)^2\big(ck^2 - ck\big)} \\[1em] \Rightarrow \dfrac{c^2k^4(ck - c)}{c^2k^2(ck^2 - ck)} \\[1em] \Rightarrow \dfrac{c^3 k^4 (k - 1)}{c^3k^3(k - 1)} \\[1em] \Rightarrow k. ⇒ b 2 ( a − b ) a 2 ( b − c ) ⇒ ( c k ) 2 ( c k 2 − c k ) ( c k 2 ) 2 ( c k − c ) ⇒ c 2 k 2 ( c k 2 − c k ) c 2 k 4 ( c k − c ) ⇒ c 3 k 3 ( k − 1 ) c 3 k 4 ( k − 1 ) ⇒ k .
Since, L.H.S. = R.H.S.
Hence, proved that a + b b + c = a 2 ( b − c ) b 2 ( a − b ) \dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)} b + c a + b = b 2 ( a − b ) a 2 ( b − c )
(ii) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
⇒ a + b + c a − b + c ⇒ c k 2 + c k + c c k 2 − c k + c ⇒ c ( k 2 + k + 1 ) c ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{a + b + c}{a - b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck + c}{ck^2 - ck + c} \\[1em] \Rightarrow \dfrac{c(k^2 + k + 1)}{c(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a − b + c a + b + c ⇒ c k 2 − c k + c c k 2 + c k + c ⇒ c ( k 2 − k + 1 ) c ( k 2 + k + 1 ) ⇒ k 2 − k + 1 k 2 + k + 1 .
Substituting values of a and b in R.H.S. of equation a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
⇒ ( a + b + c ) 2 a 2 + b 2 + c 2 ⇒ ( c k 2 + c k + c ) 2 ( c k 2 ) 2 + ( c k ) 2 + c 2 ⇒ c 2 ( k 2 + k + 1 ) 2 c 2 ( k 4 + k 2 + 1 ) ⇒ ( k 2 + k + 1 ) 2 ( k 4 + k 2 + 1 ) ⇒ ( k 2 + k + 1 ) 2 ( k 2 + k + 1 ) ( k 2 − k + 1 ) ⇒ k 2 + k + 1 k 2 − k + 1 . \Rightarrow \dfrac{(a + b + c)^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{(ck^2 + ck + c)^2}{(ck^2)^2 + (ck)^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)^2}{c^2(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^2 + k + 1)(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}. ⇒ a 2 + b 2 + c 2 ( a + b + c ) 2 ⇒ ( c k 2 ) 2 + ( c k ) 2 + c 2 ( c k 2 + c k + c ) 2 ⇒ c 2 ( k 4 + k 2 + 1 ) c 2 ( k 2 + k + 1 ) 2 ⇒ ( k 4 + k 2 + 1 ) ( k 2 + k + 1 ) 2 ⇒ ( k 2 + k + 1 ) ( k 2 − k + 1 ) ( k 2 + k + 1 ) 2 ⇒ k 2 − k + 1 k 2 + k + 1 .
Since, L.H.S. = R.H.S.
Hence, proved that a + b + c a − b + c = ( a + b + c ) 2 ( a 2 + b 2 + c 2 ) \dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})} a − b + c a + b + c = ( a 2 + b 2 + c 2 ) ( a + b + c ) 2
(iii) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of equation a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
⇒ a 2 + a b + b 2 b 2 + b c + c 2 ⇒ ( c k 2 ) 2 + ( c k 2 ) ( c k ) + ( c k ) 2 ( c k ) 2 + ( c k ) c + c 2 ⇒ c 2 k 4 + c 2 k 3 + c 2 k 2 c 2 k 2 + c 2 k + c 2 ⇒ c 2 ( k 4 + k 3 + k 2 ) c 2 ( k 2 + k + 1 ) ⇒ k 2 ( k 2 + k + 1 ) ( k 2 + k + 1 ) ⇒ k 2 . \Rightarrow \dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} \\[1em] \Rightarrow \dfrac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \\[1em] \Rightarrow \dfrac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^4 + k^3 + k^2)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow \dfrac{k^2(k^2 + k + 1)}{(k^2 + k + 1)} \\[1em] \Rightarrow k^2. ⇒ b 2 + b c + c 2 a 2 + ab + b 2 ⇒ ( c k ) 2 + ( c k ) c + c 2 ( c k 2 ) 2 + ( c k 2 ) ( c k ) + ( c k ) 2 ⇒ c 2 k 2 + c 2 k + c 2 c 2 k 4 + c 2 k 3 + c 2 k 2 ⇒ c 2 ( k 2 + k + 1 ) c 2 ( k 4 + k 3 + k 2 ) ⇒ ( k 2 + k + 1 ) k 2 ( k 2 + k + 1 ) ⇒ k 2 .
Substituting values of a and b in R.H.S. of equation a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
⇒ a c ⇒ c k 2 c ⇒ k 2 . \Rightarrow \dfrac{a}{c} \\[1em] \Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2. ⇒ c a ⇒ c c k 2 ⇒ k 2 .
Since, L.H.S. = R.H.S.
Hence, proved that a 2 + a b + b 2 b 2 + b c + c 2 = a c \dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c} b 2 + b c + c 2 a 2 + ab + b 2 = c a
(iv) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of (a + b + c)(a − b + c) = (a2 + b2 + c2 ), we get:
⇒ (a + b + c)(a − b + c)
⇒ ((ck2 ) + (ck) + c)((ck2 ) − (ck) + c)
⇒ c(k2 + k + 1). c(k2 - k + 1)
⇒ c2 (k4 + k2 + 1)
Substituting values of a and b in R.H.S. of (a + b + c)(a − b + c) = (a2 + b2 + c2 ), we get:
⇒ (a2 + b2 + c2 )
⇒ (ck2 )2 + (ck)2 + c2
⇒ (c2 k4 + c2 k2 + c2 )
⇒ c2 (k4 + k2 + 1)
Since L.H.S. = R.H.S.
Hence, proved that (a + b + c)(a − b + c) = (a2 + b2 + c2 ).
(v) Given,
⇒ a, b, c are in continued proportion
∴ a : b = b : c
⇒ a b = b c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} ⇒ b a = c b
⇒ b = ck, a = bk = (ck)k = ck2 .
Substituting values of a and b in L.H.S. of a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ), we get:
⇒ a 2 b 2 c 2 ( a − 3 + b − 3 + c − 3 ) ⇒ c 6 k 6 ( 1 c 3 k 6 + 1 c 3 k 3 + 1 c 3 ) ⇒ c 3 ( 1 + k 3 + k 6 ) . \Rightarrow a^2 b^2 c^2\big(a^{ - 3} + b^{ - 3} + c^{ - 3}\big) \\[1em] \Rightarrow c^6 k^6\Big(\dfrac{1}{c^3 k^6} + \dfrac{1}{c^3 k^3} + \dfrac{1}{c^3}\Big) \\[1em] \Rightarrow c^3\big(1 + k^3 + k^6\big). ⇒ a 2 b 2 c 2 ( a − 3 + b − 3 + c − 3 ) ⇒ c 6 k 6 ( c 3 k 6 1 + c 3 k 3 1 + c 3 1 ) ⇒ c 3 ( 1 + k 3 + k 6 ) .
Substituting values of a and b in R.H.S. of a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ), we get:
⇒ a 3 + b 3 + c 3 ⇒ ( c k 2 ) 3 + ( c k ) 3 + c 3 ⇒ c 3 k 6 + c 3 k 3 + c 3 ⇒ c 3 ( k 6 + k 3 + 1 ) . \Rightarrow a^3 + b^3 + c^3 \\[1em] \Rightarrow (ck^2)^3 + (ck)^3 + c^3 \\[1em] \Rightarrow c^3k^6 + c^3k^3 + c^3 \\[1em] \Rightarrow c^3(k^6 + k^3 + 1). ⇒ a 3 + b 3 + c 3 ⇒ ( c k 2 ) 3 + ( c k ) 3 + c 3 ⇒ c 3 k 6 + c 3 k 3 + c 3 ⇒ c 3 ( k 6 + k 3 + 1 ) .
Since L.H.S. = R.H.S.
Hence, proved that a2 b2 c2 (a−3 + b−3 + c−3 ) = (a3 + b3 + c3 ).
(vi) Since, a, b, c, d are in continued proportion.
∴ a b = b c = c d = k \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k ∴ b a = c b = d c = k
c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values in L.H.S. of the equation ad(c2 + d2 ) = c3 (b + d), we get :
L.H.S = ad(c2 + d2 )
= dk3 .(d).[(dk)2 + d2 ]
= d2 k3 .[d2 (k2 + 1)]
= d4 k3 (k2 + 1).
Substituting values in R.H.S. of the equation ad(c2 + d2 ) = c3 (b + d), we get :
R.H.S = c3 (b + d)
= (dk)3 .(dk2 + d)
= d3 k3 [d(k2 + 1)]
= d4 k3 (k2 + 1).
Since, L.H.S = R.H.S
Hence, proved that ad(c2 + d2 ) = c3 (b + d).
If x, y and z are in continued proportion, prove that :
x y 2 . z 2 + y z 2 . x 2 + z x 2 . y 2 = 1 x 3 + 1 y 3 + 1 z 3 \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3} y 2 . z 2 x + z 2 . x 2 y + x 2 . y 2 z = x 3 1 + y 3 1 + z 3 1
Answer
Given,
x, y and z are in continued proportion.
∴ x y = y z ⇒ y 2 = x z \therefore \dfrac{x}{y} = \dfrac{y}{z} \\[1em] \Rightarrow y^2 = xz ∴ y x = z y ⇒ y 2 = x z
To prove :
x y 2 . z 2 + y z 2 . x 2 + z x 2 . y 2 = 1 x 3 + 1 y 3 + 1 z 3 \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3} y 2 . z 2 x + z 2 . x 2 y + x 2 . y 2 z = x 3 1 + y 3 1 + z 3 1
Solving L.H.S.,
⇒ x y 2 . z 2 + y z 2 . x 2 + z x 2 . y 2 ⇒ x 3 + y 3 + z 3 x 2 . y 2 . z 2 ⇒ x 3 + y 3 + z 3 x 2 . x z . z 2 ⇒ x 3 + y 3 + z 3 x 3 . z 3 ⇒ x 3 x 3 . z 3 + y 3 x 3 z 3 + z 3 x 3 . z 3 ⇒ 1 z 3 + y 3 ( x z ) 3 + 1 x 3 ⇒ 1 z 3 + y 3 ( y 2 ) 3 + 1 x 3 ⇒ 1 z 3 + y 3 y 6 + 1 x 3 ⇒ 1 z 3 + 1 y 3 + 1 x 3 . \Rightarrow \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{x^3}{x^3.z^3} + \dfrac{y^3}{x^3z^3} + \dfrac{z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(xz)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(y^2)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{y^6} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{1}{y^3} + \dfrac{1}{x^3}. ⇒ y 2 . z 2 x + z 2 . x 2 y + x 2 . y 2 z ⇒ x 2 . y 2 . z 2 x 3 + y 3 + z 3 ⇒ x 2 . x z . z 2 x 3 + y 3 + z 3 ⇒ x 3 . z 3 x 3 + y 3 + z 3 ⇒ x 3 . z 3 x 3 + x 3 z 3 y 3 + x 3 . z 3 z 3 ⇒ z 3 1 + ( x z ) 3 y 3 + x 3 1 ⇒ z 3 1 + ( y 2 ) 3 y 3 + x 3 1 ⇒ z 3 1 + y 6 y 3 + x 3 1 ⇒ z 3 1 + y 3 1 + x 3 1 .
Since, L.H.S. = R.H.S.
Hence, proved that x y 2 . z 2 + y z 2 . x 2 + z x 2 . y 2 = 1 x 3 + 1 y 3 + 1 z 3 \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3} y 2 . z 2 x + z 2 . x 2 y + x 2 . y 2 z = x 3 1 + y 3 1 + z 3 1
If a, b, c, d are in continued proportion, prove that :
(b + c)(b + d) = (c + a)(c + d)
Answer
Given,
⇒ a, b, c, d are in continued proportion
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :
⇒ (b + c)(b + d)
⇒ (d k2 + d k)(d k2 + d)
⇒ d2 (k2 + k)(k2 + 1)
⇒ d2 k(k + 1)(k2 + 1).
Substituting values of a, b and c in R.H.S. of equation (b + c)(b + d) = (c + a)(c + d), we get :
⇒ (c + a)(c + d)
⇒ (dk + dk3 )(dk + d)
⇒ d2 (k + k3 )(k + 1)
⇒ d2 k(1 + k2 )(k + 1).
Since, L.H.S. = R.H.S.
Hence, (b + c)(b + d) = (c + a)(c + d).
If a, b, c, d are in continued proportion, prove that :
(a + b)(b + c) - (a + c)(b + d) = (b - c)2
Answer
Given a, b, c, d are in continued proportion.
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)2 , we get :
⇒ (dk3 + d)(dk2 + dk) - (dk3 + dk)(dk2 + d)
⇒ d(k3 + 1) dk(k + 1) - dk(k2 + 1) d(k2 + 1)
⇒ d2 k(k3 + 1)(k + 1) - d2 k(k2 + 1)(k2 + 1)
⇒ d2 k[(k4 + k3 + k + 1) - (k4 + 2k2 + 1)]
⇒ d2 k[k4 + k3 + k + 1 - k4 - 2k2 - 1]
⇒ d2 k[k3 - 2k2 + k]
⇒ d2 k2 [k2 - 2k + 1]
⇒ d2 k2 (k - 1)2
Substituting values of a, b and c in R.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)2 , we get :
⇒ (b - c)2
⇒ (dk2 - dk)2
⇒ (dk[k - 1])2
⇒ d2 k2 (k - 1)2
Since, L.H.S. = R.H.S.
Hence, (a + b)(b + c) - (a + c)(b + d) = (b - c)2 .
If a, b, c, d are in continued proportion, prove that :
(a2 − b2 )(c2 − d2 ) = (b2 − c2 )2
Answer
Given a, b, c, d are in continued proportion.
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 , we get :
⇒ (a2 - b2 )(c2 - d2 )
⇒ [(dk3 )2 - (dk2 )2 ] [(dk)2 - d2 ]
⇒ [d2 k6 - d2 k4 ] [d2 k2 - d2 ]
⇒ d4 (k6 - k4 )(k2 - 1)
⇒ d4 k4 (k2 - 1)(k2 - 1)
⇒ d4 k4 (k2 - 1)2 .
Substituting values of a, b and c in R.H.S. of (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 , we get :
⇒ (b2 - c2 )2
⇒ [(dk2 )2 - (dk)2 ]2
⇒ [d2 k4 - d2 k2 ]2
⇒ [d2 k2 (k2 - 1)]2
⇒ d4 k4 (k2 - 1)2 .
Since, L.H.S. = R.H.S.
Hence, proved that (a2 − b2 )(c2 − d2 ) = (b2 − c2 )2 .
If a, b, c, d are in continued proportion, prove that :
( a − b c + a − c c ) 2 − ( d − b c + d − c b ) 2 = ( a − d ) 2 ( 1 c 2 + 1 b 2 ) \Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) ( c a − b + c a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 + b 2 1 )
Answer
Given a, b, c, d are in continued proportion.
∴ a : b = b : c = c : d
a b = b c = c d \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} b a = c b = d c
⇒ c = dk, b = ck = (dk)k = dk2 , a = bk = (dk2 )k = dk3 .
Substituting values of a, b and c in L.H.S. of ( a − b c + a − c c ) 2 − ( d − b c + d − c b ) 2 = ( a − d ) 2 ( 1 c 2 + 1 b 2 ) \Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) ( c a − b + c a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 + b 2 1 )
⇒ ( d k 3 − d k 2 d k + d k 3 − d k 2 d k 2 ) 2 − ( d − d k 2 d k + d − d k d k 2 ) 2 ⇒ ( k ( d k 3 − d k 2 ) + d k 3 − d k d k 2 ) 2 − ( k ( d − d k 2 ) + d − d k d k 2 ) 2 ⇒ ( d k 4 − d k 3 + d k 3 − d k d k 2 ) 2 − ( k d − d k 3 + d − d k d k 2 ) 2 ⇒ ( d k 4 − d k d k 2 ) 2 − ( d − d k 3 d k 2 ) 2 ⇒ ( d k ( k 3 − 1 ) d k 2 ) 2 − ( d ( 1 − k 3 ) d k 2 ) 2 ⇒ ( d 2 k 2 ( k 3 − 1 ) 2 d 2 k 4 ) − ( d 2 ( 1 − k 3 ) 2 d 2 k 4 ) ⇒ ( ( k 3 − 1 ) 2 k 2 ) − ( ( 1 − k 3 ) 2 k 4 ) ⇒ ( k 6 + 1 − 2 k 3 k 2 ) − ( 1 + k 6 − 2 k 3 k 4 ) ⇒ ( k 2 ( k 6 + 1 − 2 k 3 ) − ( 1 + k 6 − 2 k 3 ) k 4 ) ⇒ ( k 8 + k 2 − 2 k 5 − 1 − k 6 + 2 k 3 k 4 ) . \Rightarrow \Big(\dfrac{dk^3 - dk^2}{dk} + \dfrac{dk^3 - dk^2}{dk^2}\Big)^2 - \Big(\dfrac{d - dk^2}{dk} + \dfrac{d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{k(dk^3 - dk^2) + dk^3 - dk}{dk^2}\Big)^2 - \Big(\dfrac{k(d - dk^2) + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk^4 - dk^3 + dk^3 - dk}{dk^2}\Big)^2 - \Big(\dfrac{kd - dk^3 + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk^4 - dk}{dk^2}\Big)^2 - \Big(\dfrac{d - dk^3}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk(k^3 - 1)}{dk^2}\Big)^2 - \Big(\dfrac{d(1 - k^3)}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{d^2k^2(k^3 - 1)^2}{d^2k^4}\Big) - \Big(\dfrac{d^2(1 - k^3)^2}{d^2k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{(k^3 - 1)^2}{k^2}\Big) - \Big(\dfrac{(1 - k^3)^2}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^6 + 1 - 2k^3}{k^2}\Big) - \Big(\dfrac{1 + k^6 - 2k^3}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4}\Big). ⇒ ( d k d k 3 − d k 2 + d k 2 d k 3 − d k 2 ) 2 − ( d k d − d k 2 + d k 2 d − d k ) 2 ⇒ ( d k 2 k ( d k 3 − d k 2 ) + d k 3 − d k ) 2 − ( d k 2 k ( d − d k 2 ) + d − d k ) 2 ⇒ ( d k 2 d k 4 − d k 3 + d k 3 − d k ) 2 − ( d k 2 k d − d k 3 + d − d k ) 2 ⇒ ( d k 2 d k 4 − d k ) 2 − ( d k 2 d − d k 3 ) 2 ⇒ ( d k 2 d k ( k 3 − 1 ) ) 2 − ( d k 2 d ( 1 − k 3 ) ) 2 ⇒ ( d 2 k 4 d 2 k 2 ( k 3 − 1 ) 2 ) − ( d 2 k 4 d 2 ( 1 − k 3 ) 2 ) ⇒ ( k 2 ( k 3 − 1 ) 2 ) − ( k 4 ( 1 − k 3 ) 2 ) ⇒ ( k 2 k 6 + 1 − 2 k 3 ) − ( k 4 1 + k 6 − 2 k 3 ) ⇒ ( k 4 k 2 ( k 6 + 1 − 2 k 3 ) − ( 1 + k 6 − 2 k 3 ) ) ⇒ ( k 4 k 8 + k 2 − 2 k 5 − 1 − k 6 + 2 k 3 ) .
Substituting values of a, b and c in R.H.S. of ( a − b c + a − c c ) 2 − ( d − b c + d − c b ) 2 = ( a − d ) 2 ( 1 c 2 + 1 b 2 ) \Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) ( c a − b + c a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 + b 2 1 )
= ( a − d ) 2 ( 1 c 2 + 1 b 2 ) = ( d k 3 − d ) 2 ( 1 ( d k ) 2 + 1 ( d k 2 ) 2 ) = ( d k 3 − d ) 2 ( 1 d 2 k 2 + 1 d 2 k 4 ) = d 2 d 2 k 2 ( k 3 − 1 ) 2 ( 1 − 1 k 2 ) = ( k 3 − 1 ) 2 ( k 2 − 1 ) k 4 = ( k 6 + 1 − 2 k 3 ) ( k 2 − 1 ) k 4 = k 8 − k 6 + k 2 − 1 + 2 k 3 − 2 k 5 k 4 . = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\dfrac{1}{(dk)^2} + \dfrac{1}{(dk^2)^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\dfrac{1}{d^2k^2} + \dfrac{1}{d^2k^4}\Big) \\[1em] = \dfrac{d^2}{d^2k^2}(k^3 - 1)^2 \Big(1 - \dfrac{1}{k^2} \Big) \\[1em] = \dfrac{(k^3 - 1)^2(k^2 - 1)}{k^4} \\[1em] = \dfrac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} \\[1em] = \dfrac{k^8 - k^6 + k^2 - 1 + 2k^3 - 2k^5}{k^4}. = ( a − d ) 2 ( c 2 1 + b 2 1 ) = ( d k 3 − d ) 2 ( ( d k ) 2 1 + ( d k 2 ) 2 1 ) = ( d k 3 − d ) 2 ( d 2 k 2 1 + d 2 k 4 1 ) = d 2 k 2 d 2 ( k 3 − 1 ) 2 ( 1 − k 2 1 ) = k 4 ( k 3 − 1 ) 2 ( k 2 − 1 ) = k 4 ( k 6 + 1 − 2 k 3 ) ( k 2 − 1 ) = k 4 k 8 − k 6 + k 2 − 1 + 2 k 3 − 2 k 5 .
Since, L.H.S. = R.H.S.
Hence, proved that ( a − b c + a − c c ) 2 − ( d − b c + d − c b ) 2 = ( a − d ) 2 ( 1 c 2 + 1 b 2 ) \Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) ( c a − b + c a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 + b 2 1 )
If ax = by = cz, prove that x 2 y z + y 2 z x + z 2 x y = b c a 2 + c a b 2 + a b c 2 \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} = \dfrac{bc}{a^{2}} + \dfrac{ca}{b^{2}} + \dfrac{ab}{c^{2}} yz x 2 + z x y 2 + x y z 2 = a 2 b c + b 2 c a + c 2 ab
Answer
Given,
ax = by = cz
⇒ a x a b c = b y a b c = c z a b c \Rightarrow \dfrac{ax}{abc} = \dfrac{by}{abc} = \dfrac{cz}{abc} ⇒ ab c a x = ab c b y = ab c cz
⇒ x b c = y c a = z a b = k \Rightarrow \dfrac{x}{bc} = \dfrac{y}{ca} = \dfrac{z}{ab} = k ⇒ b c x = c a y = ab z = k
Thus,
x = kbc, y = kca, z = kab
Substitute values of x, y , z in L.H.S , we get :
⇒ x 2 y z + y 2 z x + z 2 x y ⇒ ( k b c ) 2 ( k c a ) ( k a b ) + ( k c a ) 2 ( k a b ) ( k b c ) + ( k a b ) 2 ( k b c ) ( k c a ) ⇒ k 2 b 2 c 2 k 2 a 2 b c + k 2 c 2 a 2 k 2 b 2 c a + k 2 a 2 b 2 k 2 c 2 a b ⇒ b 2 c 2 a 2 b c + c 2 a 2 b 2 c a + a 2 b 2 c 2 a b ⇒ b c a 2 + c a b 2 + a b c 2 . \Rightarrow \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} \\[1em] \Rightarrow \dfrac{(kbc)^{2}}{(kca)(kab)} + \dfrac{(kca)^{2}}{(kab)(kbc)} + \dfrac{(kab)^{2}}{(kbc)(kca)} \\[1em] \Rightarrow \dfrac{k^2b^2c^2}{k^2a^2bc} + \dfrac{k^2c^2a^2}{k^2b^2ca} + \dfrac{k^2a^2b^2}{k^2c^2ab} \\[1em] \Rightarrow \dfrac{b^2c^2}{a^2bc} + \dfrac{c^2a^2}{b^2ca} + \dfrac{a^2b^2}{c^2ab} \\[1em] \Rightarrow \dfrac{bc}{a^2} + \dfrac{ca}{b^2} + \dfrac{ab}{c^2}. ⇒ yz x 2 + z x y 2 + x y z 2 ⇒ ( k c a ) ( kab ) ( kb c ) 2 + ( kab ) ( kb c ) ( k c a ) 2 + ( kb c ) ( k c a ) ( kab ) 2 ⇒ k 2 a 2 b c k 2 b 2 c 2 + k 2 b 2 c a k 2 c 2 a 2 + k 2 c 2 ab k 2 a 2 b 2 ⇒ a 2 b c b 2 c 2 + b 2 c a c 2 a 2 + c 2 ab a 2 b 2 ⇒ a 2 b c + b 2 c a + c 2 ab .
Hence, proved that x 2 y z + y 2 z x + z 2 x y = b c a 2 + c a b 2 + a b c 2 \dfrac{x^{2}}{yz} + \dfrac{y^{2}}{zx} + \dfrac{z^{2}}{xy} = \dfrac{bc}{a^{2}} + \dfrac{ca}{b^{2}} + \dfrac{ab}{c^{2}} yz x 2 + z x y 2 + x y z 2 = a 2 b c + b 2 c a + c 2 ab
If, x b + c − a = y c + a − b = z a + b − c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} b + c − a x = c + a − b y = a + b − c z x + y + z a + b + c \dfrac{x + y + z}{a + b + c} a + b + c x + y + z
Also, show that (b − c)x + (c − a)y + (a − b)z = 0.
Answer
Given,
x b + c − a = y c + a − b = z a + b − c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} b + c − a x = c + a − b y = a + b − c z
Let the common value of the given ratios be k.
x b + c − a = y c + a − b = z a + b − c = k \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k b + c − a x = c + a − b y = a + b − c z = k
Therefore,
x = k(b + c - a), y = k(c + a - b), z = k(a + b - c)
Adding x,y and z, we get:
⇒ x + y + z = k[(b + c − a) + (c + a − b) + (a + b − c)]
⇒ x + y + z = k(a + b + c)
⇒ k = x + y + z a + b + c \dfrac{x + y + z}{a + b + c} a + b + c x + y + z
Therefore,
x b + c − a = y c + a − b = z a + b − c = k = x + y + z a + b + c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k = \dfrac{x + y + z}{a + b + c} b + c − a x = c + a − b y = a + b − c z = k = a + b + c x + y + z
Given,
(b − c)x + (c − a)y + (a − b)z = 0.
Substituting value of x, y, z in L.H.S of above equation, we get :
⇒ (b − c)[k(b + c - a)] + (c − a)[ k(c + a - b)] + (a − b)[k(a + b - c)]
⇒ k[(b − c)(b + c - a) + (c − a)(c + a - b) + (a − b)(a + b - c)]
⇒ k[(b2 - c2 ) - a(b - c) + (c2 - a2 ) - b(c - a) + (a2 - b2 ) - c(a - b)]
⇒ k[b2 - c2 - ab + ac + c2 - a2 - bc + ab + a2 - b2 - ca + bc]
⇒ k[b2 - b2 + c2 - c2 + a2 - a2 - ab + ab + ac - ac - bc + bc]
⇒ k(0)
⇒ 0.
Hence, proved that x b + c − a = y c + a − b = z a + b − c = x + y + z a + b + c \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = \dfrac{x + y + z}{a + b + c} b + c − a x = c + a − b y = a + b − c z = a + b + c x + y + z
If b is the mean proportion between a and c, show that:
a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4 = c 2 a 2
Answer
Given,
Since b is the mean proportional between a and c, we have
⇒ a : b :: b : c
⇒ a b = b c \dfrac{a}{b} = \dfrac{b}{c} b a = c b
⇒ b2 = ac
Substituting value of b2 in a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4
⇒ a 4 + a 2 b 2 + ( b 2 ) 2 ( b 2 ) 2 + b 2 c 2 + c 4 ⇒ a 4 + a 2 . a c + ( a c ) 2 ( a c ) 2 + ( a c ) . c 2 + c 4 ⇒ a 2 ( a 2 + a c + c 2 ) c 2 ( a 2 + a c + c 2 ) ⇒ a 2 c 2 . \Rightarrow \dfrac{a^{4} + a^{2}b^{2} + (b^{2})^2}{(b^{2})^2 + b^{2}c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^{4} + a^{2}.ac + (ac)^2}{(ac)^2 + (ac).c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^2(a^{2} + ac + c^2)}{c^2(a^2 + ac + c^2)} \\[1em] \Rightarrow \dfrac{a^2}{c^2}. ⇒ ( b 2 ) 2 + b 2 c 2 + c 4 a 4 + a 2 b 2 + ( b 2 ) 2 ⇒ ( a c ) 2 + ( a c ) . c 2 + c 4 a 4 + a 2 . a c + ( a c ) 2 ⇒ c 2 ( a 2 + a c + c 2 ) a 2 ( a 2 + a c + c 2 ) ⇒ c 2 a 2 .
Hence, proved that a 4 + a 2 b 2 + b 4 b 4 + b 2 c 2 + c 4 = a 2 c 2 \dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}} b 4 + b 2 c 2 + c 4 a 4 + a 2 b 2 + b 4 = c 2 a 2
