Find the value of x in the following proportions :
(i) 10 : 35 = x : 42
(ii) 3 : x = 24 : 2
(iii) 2.5 : 1.5 = x : 3
(iv) x : 50 :: 3 : 2.
Answer
(i) Given, 10 : 35 = x : 42
⇒ 10 35 = x 42 ⇒ x = 10 35 × 42 ⇒ x = 420 35 ⇒ x = 12. \Rightarrow \dfrac{10}{35} = \dfrac{x}{42} \\[0.5em] \Rightarrow x = \dfrac{10}{35} \times 42 \\[0.5em] \Rightarrow x = \dfrac{420}{35} \\[0.5em] \Rightarrow x = 12. ⇒ 35 10 = 42 x ⇒ x = 35 10 × 42 ⇒ x = 35 420 ⇒ x = 12.
Hence, the value of x = 12.
(ii) Given, 3 : x = 24 : 2
⇒ 3 x = 24 2 ⇒ x = 3 24 × 2 ⇒ x = 6 24 ⇒ x = 1 4 . \Rightarrow \dfrac{3}{x} = \dfrac{24}{2} \\[0.5em] \Rightarrow x = \dfrac{3}{24} \times 2 \\[0.5em] \Rightarrow x = \dfrac{6}{24} \\[0.5em] \Rightarrow x = \dfrac{1}{4}. ⇒ x 3 = 2 24 ⇒ x = 24 3 × 2 ⇒ x = 24 6 ⇒ x = 4 1 .
Hence, the value of x = 1 4 \dfrac{1}{4} 4 1
(iii) Given, 2.5 : 1.5 = x : 3
⇒ 2.5 1.5 = x 3 ⇒ x = 2.5 1.5 × 3 ⇒ x = 7.5 1.5 ⇒ x = 5. \Rightarrow \dfrac{2.5}{1.5} = \dfrac{x}{3} \\[0.5em] \Rightarrow x = \dfrac{2.5}{1.5} \times 3 \\[0.5em] \Rightarrow x = \dfrac{7.5}{1.5} \\[0.5em] \Rightarrow x = 5. ⇒ 1.5 2.5 = 3 x ⇒ x = 1.5 2.5 × 3 ⇒ x = 1.5 7.5 ⇒ x = 5.
Hence, the value of x = 5.
(iv) Given, x : 50 :: 3 : 2
⇒ x 50 = 3 2 ⇒ x = 3 2 × 50 ⇒ x = 150 2 ⇒ x = 75. \Rightarrow \dfrac{x}{50} = \dfrac{3}{2} \\[0.5em] \Rightarrow x = \dfrac{3}{2} \times 50 \\[0.5em] \Rightarrow x = \dfrac{150}{2} \\[0.5em] \Rightarrow x = 75. ⇒ 50 x = 2 3 ⇒ x = 2 3 × 50 ⇒ x = 2 150 ⇒ x = 75.
Hence, the value of x = 75.
Find the fourth proportional to :
(i) 3, 12, 15
(ii) 1 3 , 1 4 , 1 5 \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5} 3 1 , 4 1 , 5 1
(iii) 1.5, 2.5, 4.5
(iv) 9.6 kg, 7.2 kg, 28.8 kg.
Answer
(i) Let the fourth proportion be x.
Then 3, 12, 15 and x are in proportion.
⇒ 3 : 12 : : 15 : x ⇒ 3 12 = 15 x ⇒ x = 15 3 × 12 ⇒ x = 180 3 ⇒ x = 60. \Rightarrow 3 : 12 :: 15 : x \\[0.5em] \Rightarrow \dfrac{3}{12} = \dfrac{15}{x} \\[0.5em] \Rightarrow x = \dfrac{15}{3} \times 12 \\[0.5em] \Rightarrow x = \dfrac{180}{3} \\[0.5em] \Rightarrow x = 60. ⇒ 3 : 12 :: 15 : x ⇒ 12 3 = x 15 ⇒ x = 3 15 × 12 ⇒ x = 3 180 ⇒ x = 60.
Hence, the fourth proportion is 60.
(ii) Let the fourth proportion be x.
Then 1 3 , 1 4 , 1 5 \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5} 3 1 , 4 1 , 5 1
⇒ 1 3 : 1 4 : : 1 5 : x ⇒ 1 3 1 4 = 1 5 x ⇒ x = 1 5 1 3 × 1 4 ⇒ x = 3 5 × 1 4 ⇒ x = 3 20 . \Rightarrow \dfrac{1}{3} : \dfrac{1}{4} :: \dfrac{1}{5} : x \\[0.5em] \Rightarrow \dfrac{\dfrac{1}{3}}{\dfrac{1}{4}} = \dfrac{\dfrac{1}{5}}{x} \\[0.5em] \Rightarrow x = \dfrac{\dfrac{1}{5}}{\dfrac{1}{3}} \times \dfrac{1}{4} \\[0.5em] \Rightarrow x = \dfrac{3}{5} \times \dfrac{1}{4} \\[0.5em] \Rightarrow x = \dfrac{3}{20}. ⇒ 3 1 : 4 1 :: 5 1 : x ⇒ 4 1 3 1 = x 5 1 ⇒ x = 3 1 5 1 × 4 1 ⇒ x = 5 3 × 4 1 ⇒ x = 20 3 .
Hence, the fourth proportion is 3 20 \dfrac{3}{20} 20 3
(iii) Let the fourth proportion be x.
Then 1.5, 2.5, 4.5 and x are in proportion.
⇒ 1.5 : 2.5 : : 4.5 : x ⇒ 1.5 2.5 = 4.5 x ⇒ x = 4.5 1.5 × 2.5 ⇒ x = 3 × 2.5 ⇒ x = 7.5. \Rightarrow 1.5 : 2.5 :: 4.5 : x \\[0.5em] \Rightarrow \dfrac{1.5}{2.5} = \dfrac{4.5}{x} \\[0.5em] \Rightarrow x = \dfrac{4.5}{1.5} \times 2.5 \\[0.5em] \Rightarrow x = 3 \times 2.5 \\[0.5em] \Rightarrow x = 7.5. ⇒ 1.5 : 2.5 :: 4.5 : x ⇒ 2.5 1.5 = x 4.5 ⇒ x = 1.5 4.5 × 2.5 ⇒ x = 3 × 2.5 ⇒ x = 7.5.
Hence, the fourth proportion is 7.5.
(iv) Let the fourth proportion be x.
Then 9.6 kg, 7.2 kg, 28.8 kg and x are in proportion.
⇒ 9.6 : 7.2 : : 28.8 : x ⇒ 9.6 7.2 = 28.8 x ⇒ x = 28.8 9.6 × 7.2 ⇒ x = 3 × 7.2 ⇒ x = 21.6. \Rightarrow 9.6 : 7.2 :: 28.8 : x \\[0.5em] \Rightarrow \dfrac{9.6}{7.2} = \dfrac{28.8}{x} \\[0.5em] \Rightarrow x = \dfrac{28.8}{9.6} \times 7.2 \\[0.5em] \Rightarrow x = 3 \times 7.2 \\[0.5em] \Rightarrow x = 21.6. ⇒ 9.6 : 7.2 :: 28.8 : x ⇒ 7.2 9.6 = x 28.8 ⇒ x = 9.6 28.8 × 7.2 ⇒ x = 3 × 7.2 ⇒ x = 21.6.
Hence, the fourth proportion is 21.6 kg.
Find the third proportional to :
(i) 5, 10
(ii) 0.24, 0.6
(iii) ₹3, ₹12
(iv) 5 1 4 5\dfrac{1}{4} 5 4 1
Answer
(i) Let the third proportion be x, then 5, 10, x are in continued proportion.
⇒ 5 10 = 10 x ⇒ x = 10 5 × 10 ⇒ x = 100 5 ⇒ x = 20. \Rightarrow \dfrac{5}{10} = \dfrac{10}{x} \\[0.5em] \Rightarrow x = \dfrac{10}{5} \times 10 \\[0.5em] \Rightarrow x = \dfrac{100}{5} \\[0.5em] \Rightarrow x = 20. ⇒ 10 5 = x 10 ⇒ x = 5 10 × 10 ⇒ x = 5 100 ⇒ x = 20.
Hence, the third proportion is 20.
(ii) Let the third proportion be x, then 0.24, 0.6, x are in continued proportion.
⇒ 0.24 0.6 = 0.6 x ⇒ x = 0.6 0.24 × 0.6 ⇒ x = 0.36 0.24 ⇒ x = 1.5. \Rightarrow \dfrac{0.24}{0.6} = \dfrac{0.6}{x} \\[0.5em] \Rightarrow x = \dfrac{0.6}{0.24} \times 0.6 \\[0.5em] \Rightarrow x = \dfrac{0.36}{0.24} \\[0.5em] \Rightarrow x = 1.5. ⇒ 0.6 0.24 = x 0.6 ⇒ x = 0.24 0.6 × 0.6 ⇒ x = 0.24 0.36 ⇒ x = 1.5.
Hence, the third proportion is 1.5.
(iii) Let the third proportion be x, then 3, 12, x are in continued proportion.
⇒ 3 12 = 12 x ⇒ x = 12 3 × 12 ⇒ x = 144 3 ⇒ x = 48. \Rightarrow \dfrac{3}{12} = \dfrac{12}{x} \\[0.5em] \Rightarrow x = \dfrac{12}{3} \times 12 \\[0.5em] \Rightarrow x = \dfrac{144}{3} \\[0.5em] \Rightarrow x = 48. ⇒ 12 3 = x 12 ⇒ x = 3 12 × 12 ⇒ x = 3 144 ⇒ x = 48.
Hence, the third proportion is ₹ 48.
(iv) Let the third proportion be x, then 5 1 4 5\dfrac{1}{4} 5 4 1
5 1 4 = 21 4 ⇒ 21 4 7 = 7 x ⇒ x = 7 21 4 × 7 ⇒ x = 7 × 4 × 7 21 ⇒ x = 28 3 ⇒ x = 9 1 3 . 5\dfrac{1}{4} = \dfrac{21}{4} \\[0.5em] \Rightarrow \dfrac{\dfrac{21}{4}}{7} = \dfrac{7}{x} \\[0.5em] \Rightarrow x = \dfrac{7}{\dfrac{21}{4}} \times 7 \\[0.5em] \Rightarrow x = \dfrac{7 \times 4 \times 7}{21} \\[0.5em] \Rightarrow x = \dfrac{28}{3} \\[0.5em] \Rightarrow x = 9\dfrac{1}{3}. 5 4 1 = 4 21 ⇒ 7 4 21 = x 7 ⇒ x = 4 21 7 × 7 ⇒ x = 21 7 × 4 × 7 ⇒ x = 3 28 ⇒ x = 9 3 1 .
Hence, the third proportion is 9 1 3 9\dfrac{1}{3} 9 3 1
Find the mean proportion of :
(i) 5 and 80
(ii) 1 12 \dfrac{1}{12} 12 1 1 75 \dfrac{1}{75} 75 1
(iii) 8.1 and 2.5
(iv) (a - b) and (a3 - a2 b), a > b.
Answer
(i) Let the mean proportion be x.
∴ 5 x = x 80 ⇒ x 2 = 5 × 80 ⇒ x 2 = 400 ⇒ x = 400 ⇒ x = 20. \therefore \dfrac{5}{x} = \dfrac{x}{80} \\[0.5em] \Rightarrow x^2 = 5 \times 80 \\[0.5em] \Rightarrow x^2 = 400 \\[0.5em] \Rightarrow x = \sqrt{400} \\[0.5em] \Rightarrow x = 20. ∴ x 5 = 80 x ⇒ x 2 = 5 × 80 ⇒ x 2 = 400 ⇒ x = 400 ⇒ x = 20.
Hence, the mean proportion is 20.
(ii) Let the mean proportion be x.
∴ 1 12 x = x 1 75 ⇒ x 2 = 1 12 × 1 75 ⇒ x 2 = 1 900 ⇒ x = 1 900 ⇒ x = 1 30 \therefore \dfrac{\dfrac{1}{12}}{x} = \dfrac{x}{\dfrac{1}{75}} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{12} \times \dfrac{1}{75} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{900} \\[0.5em] \Rightarrow x = \sqrt{\dfrac{1}{900}} \\[0.5em] \Rightarrow x = \dfrac{1}{30} ∴ x 12 1 = 75 1 x ⇒ x 2 = 12 1 × 75 1 ⇒ x 2 = 900 1 ⇒ x = 900 1 ⇒ x = 30 1
Hence, the mean proportion is 1 30 \dfrac{1}{30} 30 1
(iii) Let the mean proportion be x.
∴ 8.1 x = x 2.5 ⇒ x 2 = 8.1 × 2.5 ⇒ x 2 = 20.25 ⇒ x = 20.25 ⇒ x = 4.5. \therefore \dfrac{8.1}{x} = \dfrac{x}{2.5} \\[0.5em] \Rightarrow x^2 = 8.1 \times 2.5 \\[0.5em] \Rightarrow x^2 = 20.25 \\[0.5em] \Rightarrow x = \sqrt{20.25} \\[0.5em] \Rightarrow x = 4.5. ∴ x 8.1 = 2.5 x ⇒ x 2 = 8.1 × 2.5 ⇒ x 2 = 20.25 ⇒ x = 20.25 ⇒ x = 4.5.
Hence, the mean proportion is 4.5.
(iv) Let the mean proportion be x.
∴ ( a − b ) x = x ( a 3 − a 2 b ) ⇒ x 2 = ( a − b ) × ( a 3 − a 2 b ) ⇒ x 2 = ( a 4 − a 3 b − a 3 b + a 2 b 2 ) ⇒ x 2 = ( a 4 − 2 a 3 b + a 2 b 2 ) ⇒ x 2 = a 2 ( a 2 − 2 a b + b 2 ) ⇒ x 2 = a 2 ( a − b ) 2 ⇒ x = ( a 2 ( a − b ) 2 ) ⇒ x = a ( a − b ) . \therefore \dfrac{(a - b)}{x} = \dfrac{x}{(a^3 - a^2b)} \\[0.5em] \Rightarrow x^2 = (a - b) \times (a^3 - a^2b) \\[0.5em] \Rightarrow x^2 = (a^4 - a^3b - a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = (a^4 - 2a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = a^2(a^2 - 2ab + b^2) \\[0.5em] \Rightarrow x^2 = a^2(a - b)^2 \\[0.5em] \Rightarrow x = \sqrt{(a^2(a - b)^2)} \\[0.5em] \Rightarrow x = a(a - b). ∴ x ( a − b ) = ( a 3 − a 2 b ) x ⇒ x 2 = ( a − b ) × ( a 3 − a 2 b ) ⇒ x 2 = ( a 4 − a 3 b − a 3 b + a 2 b 2 ) ⇒ x 2 = ( a 4 − 2 a 3 b + a 2 b 2 ) ⇒ x 2 = a 2 ( a 2 − 2 ab + b 2 ) ⇒ x 2 = a 2 ( a − b ) 2 ⇒ x = ( a 2 ( a − b ) 2 ) ⇒ x = a ( a − b ) .
Hence, the mean proportion is a(a - b).
If a, 12, 16 and b are in continued proportion, find a and b.
Answer
Given, a, 12, 16 and b are in continued proportion.
∴ a 12 = 12 16 = 16 b ⇒ a 12 = 12 16 and 12 16 = 16 b ⇒ a = 12 16 × 12 and b = 16 12 × 16 ⇒ a = 144 16 and b = 256 12 ⇒ a = 9 and b = 64 3 . \therefore \dfrac{a}{12} = \dfrac{12}{16} = \dfrac{16}{b} \\[1em] \Rightarrow \dfrac{a}{12} = \dfrac{12}{16} \text{ and } \dfrac{12}{16} = \dfrac{16}{b} \\[1em] \Rightarrow a = \dfrac{12}{16} \times 12 \text{ and } b = \dfrac{16}{12} \times 16 \\[1em] \Rightarrow a = \dfrac{144}{16} \text{ and } b = \dfrac{256}{12} \\[1em] \Rightarrow a = 9 \text{ and } b = \dfrac{64}{3}. ∴ 12 a = 16 12 = b 16 ⇒ 12 a = 16 12 and 16 12 = b 16 ⇒ a = 16 12 × 12 and b = 12 16 × 16 ⇒ a = 16 144 and b = 12 256 ⇒ a = 9 and b = 3 64 .
Hence, the value of a = 9 and b = 64 3 \dfrac{64}{3} 3 64
What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Answer
Let the number to be added be x. So, new numbers are 5 + x, 11 + x, 19 + x, 37 + x
Since, these numbers are in proportion,
∴ 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 − x 2 + 42 x − 30 x + 185 − 209 = 0 ⇒ 12 x − 24 = 0 ⇒ 12 x = 24 ⇒ x = 24 12 = 2. \therefore \dfrac{5 + x}{11 + x} = \dfrac{19 + x}{37 + x} \\[0.5em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[0.5em] (185 + 5x + 37x + x^2) = 209 + 19x + 11x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 42x - 30x + 185 - 209 = 0 \\[0.5em] \Rightarrow 12x - 24 = 0 \\[0.5em] \Rightarrow 12x = 24 \\[0.5em] \Rightarrow x = \dfrac{24}{12} = 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 − x 2 + 42 x − 30 x + 185 − 209 = 0 ⇒ 12 x − 24 = 0 ⇒ 12 x = 24 ⇒ x = 12 24 = 2.
Hence, the number that must be added to make the numbers in proportion is 2.
What numbers should be subtracted from each of the numbers 23, 30, 57 and 78 so that remainders are in proportion?
Answer
Let the number to be subtracted be x. So, new numbers are 23 - x, 30 - x, 57 - x, 78 - x
Since, these numbers are in proportion,
∴ 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 − x 2 − 101 x + 87 x + 1794 − 1710 = 0 ⇒ − 14 x + 84 = 0 ⇒ 14 x = 84 ⇒ x = 6. \therefore \dfrac{23 - x}{30 - x} = \dfrac{57 - x}{78 - x} \\[0.5em] \Rightarrow (23 - x)(78 - x) = (57 - x)(30 - x) \\[0.5em] \Rightarrow (1794 - 23x - 78x + x^2) = (1710 - 57x - 30x + x^2) \\[0.5em] \Rightarrow x^2 - x^2 - 101x + 87x + 1794 - 1710 = 0 \\[0.5em] \Rightarrow -14x + 84 = 0 \\[0.5em] \Rightarrow 14x = 84 \\[0.5em] \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 − x 2 − 101 x + 87 x + 1794 − 1710 = 0 ⇒ − 14 x + 84 = 0 ⇒ 14 x = 84 ⇒ x = 6.
Hence, the number that must be subtracted to make the numbers in proportion is 6.
If k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion, find k.
Answer
Since, k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion,
∴ 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 ⇒ 2 k 2 − 3 k 2 + 3 k + k − 9 + 14 = 0 ⇒ − k 2 + 4 k + 5 = 0 \therefore \dfrac{k + 3}{k + 2} = \dfrac{3k - 7}{2k - 3} \\[0.5em] \Rightarrow (k + 3)(2k - 3) = (3k - 7)(k + 2) \\[0.5em] \Rightarrow 2k^2 - 3k + 6k - 9 = 3k^2 + 6k - 7k - 14 \\[0.5em] \Rightarrow 2k^2 + 3k - 9 = 3k^2 - k - 14 \\[0.5em] \Rightarrow 2k^2 - 3k^2 + 3k + k - 9 + 14 = 0 \\[0.5em] \Rightarrow -k^2 + 4k + 5 = 0 ∴ 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 ⇒ 2 k 2 − 3 k 2 + 3 k + k − 9 + 14 = 0 ⇒ − k 2 + 4 k + 5 = 0
Multiplying equation by -1,
⇒ k 2 − 4 k − 5 = 0 ⇒ k 2 − 5 k + k − 5 = 0 ⇒ k ( k − 5 ) + 1 ( k − 5 ) = 0 ⇒ ( k + 1 ) ( k − 5 ) = 0 ⇒ k + 1 = 0 or k − 5 = 0 ⇒ k = − 1 or k = 5. \Rightarrow k^2 - 4k - 5 = 0 \\[0.5em] \Rightarrow k^2 - 5k + k - 5 = 0 \\[0.5em] \Rightarrow k(k - 5) + 1(k - 5) = 0 \\[0.5em] \Rightarrow (k + 1)(k - 5) = 0 \\[0.5em] \Rightarrow k + 1 = 0 \text{ or } k - 5 = 0 \\[0.5em] \Rightarrow k = -1 \text{ or } k = 5. ⇒ k 2 − 4 k − 5 = 0 ⇒ k 2 − 5 k + k − 5 = 0 ⇒ k ( k − 5 ) + 1 ( k − 5 ) = 0 ⇒ ( k + 1 ) ( k − 5 ) = 0 ⇒ k + 1 = 0 or k − 5 = 0 ⇒ k = − 1 or k = 5.
Hence, the value of k is -1 and 5.
If (x + 5) is the mean proportion between x + 2 and x + 9, find the value of x.
Answer
Since, (x + 5) is the mean proportion between x + 2 and x + 9,
∴ x + 2 x + 5 = x + 5 x + 9 ⇒ ( x + 5 ) 2 = ( x + 2 ) ( x + 9 ) ⇒ ( x 2 + 25 + 10 x ) = ( x 2 + 9 x + 2 x + 18 ) ⇒ x 2 − x 2 + 10 x − 11 x + 25 − 18 = 0 ⇒ − x + 7 = 0 ⇒ x = 7. \therefore \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} \\[0.5em] \Rightarrow (x + 5)^2 = (x + 2)(x + 9) \\[0.5em] \Rightarrow (x^2 + 25 + 10x) = (x^2 + 9x + 2x + 18) \\[0.5em] \Rightarrow x^2 - x^2 + 10x - 11x + 25 - 18 = 0 \\[0.5em] \Rightarrow -x + 7 = 0 \\[0.5em] \Rightarrow x = 7. ∴ x + 5 x + 2 = x + 9 x + 5 ⇒ ( x + 5 ) 2 = ( x + 2 ) ( x + 9 ) ⇒ ( x 2 + 25 + 10 x ) = ( x 2 + 9 x + 2 x + 18 ) ⇒ x 2 − x 2 + 10 x − 11 x + 25 − 18 = 0 ⇒ − x + 7 = 0 ⇒ x = 7.
Hence, the value of x is 7.
What numbers must be added to each of the numbers 16, 26 and 40 so that the resulting numbers must be in continued proportion?
Answer
Let the number to be added to each number be x. So, the new numbers are 16 + x, 26 + x, 40 + x.
Since, new numbers are in continued proportion,
∴ 16 + x 26 + x = 26 + x 40 + x ⇒ ( 16 + x ) ( 40 + x ) = ( 26 + x ) ( 26 + x ) ⇒ 640 + 16 x + 40 x + x 2 = 676 + 26 x + 26 x + x 2 ⇒ x 2 − x 2 + 56 x − 52 x + 640 − 676 = 0 ⇒ 4 x − 36 = 0 ⇒ 4 x = 36 ⇒ x = 9. \therefore \dfrac{16 + x}{26 + x} = \dfrac{26 + x}{40 + x} \\[0.5em] \Rightarrow (16 + x)(40 + x) = (26 + x)(26 + x) \\[0.5em] \Rightarrow 640 + 16x + 40x + x^2 = 676 + 26x + 26x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 56x - 52x + 640 - 676 = 0 \\[0.5em] \Rightarrow 4x - 36 = 0 \\[0.5em] \Rightarrow 4x = 36 \Rightarrow x = 9. ∴ 26 + x 16 + x = 40 + x 26 + x ⇒ ( 16 + x ) ( 40 + x ) = ( 26 + x ) ( 26 + x ) ⇒ 640 + 16 x + 40 x + x 2 = 676 + 26 x + 26 x + x 2 ⇒ x 2 − x 2 + 56 x − 52 x + 640 − 676 = 0 ⇒ 4 x − 36 = 0 ⇒ 4 x = 36 ⇒ x = 9.
Hence, the number to be added to each number is 9.
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Answer
Let the two numbers be x and y.
Given, 28 is the mean proportional between x and y.
∴ x 28 = 28 y ⇒ x y = 28 × 28 ⇒ x y = 784 ⇒ x = 784 y [....Eq 1] \therefore \dfrac{x}{28} = \dfrac{28}{y} \\[0.5em] \Rightarrow xy = 28 \times 28 \\[0.5em] \Rightarrow xy = 784 \\[0.5em] \Rightarrow x = \dfrac{784}{y} \qquad \text{[....Eq 1]} ∴ 28 x = y 28 ⇒ x y = 28 × 28 ⇒ x y = 784 ⇒ x = y 784 [....Eq 1]
Given, 224 is the third proportional to numbers.
∴ x y = y 224 ⇒ y 2 = 224 x \therefore \dfrac{x}{y} = \dfrac{y}{224} \\[0.5em] \Rightarrow y^2 = 224x \\[0.5em] ∴ y x = 224 y ⇒ y 2 = 224 x
Putting value of x from equation 1 above:
⇒ y 2 = 224 ( 784 y ) ⇒ y 3 = 175616 ⇒ y = 175616 3 ⇒ y = 56 3 3 ⇒ y = 56 and x = 784 56 = 14. \Rightarrow y^2 = 224\Big(\dfrac{784}{y}\Big) \\[0.5em] \Rightarrow y^3 = 175616 \\[0.5em] \Rightarrow y = \sqrt[3]{175616} \\[0.5em] \Rightarrow y = \sqrt[3]{56^3} \\[0.5em] \Rightarrow y = 56 \\[0.5em] \text{ and } x = \dfrac{784}{56} = 14. ⇒ y 2 = 224 ( y 784 ) ⇒ y 3 = 175616 ⇒ y = 3 175616 ⇒ y = 3 5 6 3 ⇒ y = 56 and x = 56 784 = 14.
Hence, the two numbers are 14 and 56.
If b is the mean proportional between a and c , prove that a, c, a2 + b2 and b2 + c2 are proportional.
Answer
Given, b is the mean proportional between a and c then,
b2 = ac. [....Eq 1]
If a, c, a2 + b2 and b2 + c2 are proportional then,
⇒ a c = a 2 + b 2 b 2 + c 2 ⇒ a ( b 2 + c 2 ) = c ( a 2 + b 2 ) \Rightarrow \dfrac{a}{c} = \dfrac{a^2 + b^2}{b^2 + c^2} \\[0.5em] \Rightarrow a(b^2 + c^2) = c(a^2 + b^2) \\[0.5em] ⇒ c a = b 2 + c 2 a 2 + b 2 ⇒ a ( b 2 + c 2 ) = c ( a 2 + b 2 )
Solving L.H.S first
a(b2 + c2 )2 ) [Putting value of b2 from Eq 1]
Solving R.H.S
c(a2 + b2 )2 + ac) [Putting value of b2 from Eq 1]
Since, L.H.S. = R.H.S = ac(a + c), hence the numbers, a, c, a2 + b2 and b2 + c2 are in proportion.
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a2 + b2 ) and (b2 + c2 ).
Answer
Given, b is the mean proportional between a and c then,
b2 = ac. [....Eq 1]
For (ab + bc) to be the mean proportional between (a2 + b2 ) and (b2 + c2 ) following condition must be satisfied,
(ab + bc)2 = (a2 + b2 )(b2 + c2 )
Solving L.H.S. first,
⇒ ( a b + b c ) 2 = a 2 b 2 + 2 a b 2 c + b 2 c 2 \Rightarrow (ab + bc)^2 \\[0.5em] = a^2b^2 + 2ab^2c + b^2c^2 \\[0.5em] ⇒ ( ab + b c ) 2 = a 2 b 2 + 2 a b 2 c + b 2 c 2
Putting value of b2 from equation 1:
= a 2 ( a c ) + 2 a c ( a c ) + c 2 ( a c ) = a 3 c + 2 a 2 c 2 + a c 3 = a c ( a 2 + c 2 + 2 a c ) = a c ( a + c ) 2 = a^2(ac) + 2ac(ac) + c^2(ac) \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + c^2 + 2ac) \\[0.5em] = ac(a + c)^2 = a 2 ( a c ) + 2 a c ( a c ) + c 2 ( a c ) = a 3 c + 2 a 2 c 2 + a c 3 = a c ( a 2 + c 2 + 2 a c ) = a c ( a + c ) 2
Now, solving R.H.S. ,
⇒ ( a 2 + b 2 ) ( b 2 + c 2 ) = ( a 2 b 2 + a 2 c 2 + b 4 + b 2 c 2 ) \Rightarrow (a^2 + b^2)(b^2 + c^2) \\[0.5em] = (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \\[0.5em] ⇒ ( a 2 + b 2 ) ( b 2 + c 2 ) = ( a 2 b 2 + a 2 c 2 + b 4 + b 2 c 2 )
Putting value of b2 from equation 1:
= ( a 2 ( a c ) + a 2 c 2 + ( a c ) 2 + ( a c ) ( c 2 ) = a 3 c + a 2 c 2 + a 2 c 2 + a c 3 = a 3 c + 2 a 2 c 2 + a c 3 = a c ( a 2 + 2 a c + c 2 ) = a c ( a + c ) 2 = (a^2(ac) + a^2c^2 + (ac)^2 + (ac)(c^2) \\[0.5em] = a^3c + a^2c^2 + a^2c^2 + ac^3 \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + 2ac + c^2) \\[0.5em] = ac(a + c)^2 = ( a 2 ( a c ) + a 2 c 2 + ( a c ) 2 + ( a c ) ( c 2 ) = a 3 c + a 2 c 2 + a 2 c 2 + a c 3 = a 3 c + 2 a 2 c 2 + a c 3 = a c ( a 2 + 2 a c + c 2 ) = a c ( a + c ) 2
Since, L.H.S. = R.H.S. = ac(a + c)2 hence, (ab + bc) is the mean proportional between (a2 + b2 ) and (b2 + c2 ).
If y is the mean proportional between x and z, prove that
xyz(x + y + z)3 = (xy + yz + zx)3
Answer
Given, y is the mean proportional between x and z then,
y2 = xz [....Eq 1]
Given, xyz(x + y + z)3 = (xy + yz + zx)3
Solving L.H.S. first,
xyz(x + y + z)3 3 2 from equation 1:2 .y(x + y + z)3 3 (x + y + z)3 3 2 + yz)3 3 = R.H.S.
L.H.S. = R.H.S. , hence proved, xyz(x + y + z)3 = (xy + yz + zx)3 .
If a + c = mb and 1 b + 1 d = m c \dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c} b 1 + d 1 = c m
Answer
Given,
a + c = mb and 1 b + 1 d = m c \dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c} b 1 + d 1 = c m
Solving, a + c = mb
Dividing the equation by b,
⇒ a b + c b = m \Rightarrow \dfrac{a}{b} + \dfrac{c}{b} = m ⇒ b a + b c = m
Now solving,
1 b + 1 d = m c \dfrac{1}{b} + \dfrac{1}{d} = \dfrac{m}{c} b 1 + d 1 = c m
Multiplying the equation by c,
⇒ c b + c d = m \Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = m ⇒ b c + d c = m
Putting the value of m from Equation 1,
⇒ c b + c d = a b + c b ⇒ a b + c b = c b + c d ⇒ a b = c d \Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = \dfrac{a}{b} + \dfrac{c}{b} \\[0.5em] \Rightarrow \dfrac{a}{b} + \bcancel{\dfrac{c}{b}} = \bcancel{\dfrac{c}{b}} + \dfrac{c}{d} \\[0.5em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} ⇒ b c + d c = b a + b c ⇒ b a + b c = b c + d c ⇒ b a = d c
Since, a b = c d \dfrac{a}{b} = \dfrac{c}{d} b a = d c
If x a = y b = z c , \dfrac{x}{a} =\dfrac{y}{b} = \dfrac{z}{c}, a x = b y = c z ,
(i) x 3 a 2 + y 3 b 2 + z 3 c 2 = ( x + y + z ) 3 ( a + b + c ) 2 (ii) ( a 2 x 2 + b 2 y 2 + c 2 z 2 a 3 x + b 3 y + c 3 z ) 3 = x y z a b c (iii) 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. \begin{array}{ll} \text{(i)} & \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\ \text{(ii)} & \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \dfrac{xyz}{abc} \\ \text{(iii)} & \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} \\ & + \dfrac{cz - ax}{(c + a)(z - x)} = 3. \end{array} (i) (ii) (iii) a 2 x 3 + b 2 y 3 + c 2 z 3 = ( a + b + c ) 2 ( x + y + z ) 3 ( a 3 x + b 3 y + c 3 z a 2 x 2 + b 2 y 2 + c 2 z 2 ) 3 = ab c x yz ( 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) Let x a = y b = z c = k \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k a x = b y = c z = k
∴ x = ak, y = bk, z = ck.
L.H.S. = x 3 a 2 + y 3 b 2 + z 3 c 2 = a 3 k 3 a 2 + b 3 k 3 b 2 + c 3 k 3 c 2 = a k 3 + b k 3 + c k 3 = k 3 ( a + b + c ) . \text{L.H.S.} = \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[0.5em] = \dfrac{a^3k^3}{a^2} + \dfrac{b^3k^3}{b^2} + \dfrac{c^3k^3}{c^2} \\[0.5em] = ak^3 + bk^3 + ck^3 \\[0.5em] =k^3(a + b + c). L.H.S. = a 2 x 3 + b 2 y 3 + c 2 z 3 = a 2 a 3 k 3 + b 2 b 3 k 3 + c 2 c 3 k 3 = a k 3 + b k 3 + c k 3 = k 3 ( a + b + c ) .
R.H.S. = ( x + y + z ) 3 ( a + b + c ) 2 = ( a k + b k + c k ) 3 ( a + b + c ) 2 = k 3 ( a + b + c ) 3 ( a + b + c ) 2 = k 3 ( a + b + c ) . \text{R.H.S.} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{(ak + bk + ck)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{k^3(a + b + c)^3}{(a + b + c)^2} \\[0.5em] = k^3(a + b + c). R.H.S. = ( a + b + c ) 2 ( x + y + z ) 3 = ( a + b + c ) 2 ( ak + bk + c k ) 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
(ii) Let x a = y b = z c = k \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k a x = b y = c z = k
∴ x = ak, y = bk, z = ck.
L.H.S. = ( a 2 x 2 + b 2 y 2 + c 2 z 2 a 3 x + b 3 y + c 3 ) 3 = ( a 2 ( a k ) x + b 2 ( b k ) y + c 2 ( c k ) z a 3 x + b 3 y + c 3 z ) 3 = ( a 3 x k + b 3 y k + c 3 z k a 3 x + b 3 y + c 3 z ) 3 = k 3 ( a 3 x + b 3 y + c 3 z ) 3 ( a 3 x + b 3 y + c 3 z ) 3 = k 3 = k × k × k = x a × y b × z c = x y z a b c = R.H.S. \text{L.H.S.} = \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3}\Big)^3 \\[1em] = \Big(\dfrac{a^2(ak)x + b^2(bk)y + c^2(ck)z}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \Big(\dfrac{a^3xk + b^3yk + c^3zk}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \dfrac{k^3(a^3x + b^3y + c^3z)^3}{(a^3x + b^3y + c^3z)^3} \\[1em] = k^3 \\[1em] = k \times k \times k \\[1em] = \dfrac{x}{a} \times \dfrac{y}{b} \times \dfrac{z}{c} \\[1em] = \dfrac{xyz}{abc} = \text{R.H.S.} L.H.S. = ( a 3 x + b 3 y + c 3 a 2 x 2 + b 2 y 2 + c 2 z 2 ) 3 = ( a 3 x + b 3 y + c 3 z a 2 ( ak ) x + b 2 ( bk ) y + c 2 ( c k ) z ) 3 = ( a 3 x + b 3 y + c 3 z a 3 x k + b 3 y k + c 3 z k ) 3 = ( a 3 x + b 3 y + c 3 z ) 3 k 3 ( a 3 x + b 3 y + c 3 z ) 3 = k 3 = k × k × k = a x × b y × c z = ab c x yz = R.H.S.
Since, L.H.S. = R.H.S.
Hence proved, that ( a 2 x 2 + b 2 y 2 + c 2 z 2 a 3 x + b 3 y + c 3 z ) 3 = x y z a b c \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \dfrac{xyz}{abc} ( a 3 x + b 3 y + c 3 z a 2 x 2 + b 2 y 2 + c 2 z 2 ) 3 = ab c x yz
(iii) Let x a = y b = z c = k \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k a x = b y = c z = k
∴ x = ak, y = bk, z = ck.
L.H.S. = a x − b y ( a + b ) ( x − y ) + b y − c z ( b + c ) ( y − z ) + c z − a x ( c + a ) ( z − x ) = a ( a k ) − b ( b k ) ( a + b ) ( ( a k ) − ( b k ) ) + b ( b k ) − c ( c k ) ( b + c ) ( b k − c k ) + c ( c k ) − a ( a k ) ( c + a ) ( c k − a k ) = a 2 k − b 2 k k ( a + b ) ( a − b ) + b 2 k − c 2 k k ( b + c ) ( b − c ) + c 2 k − a 2 k k ( c + a ) ( c − a ) = k ( a 2 − b 2 ) k ( a 2 − b 2 ) + k ( b 2 − c 2 ) k ( b 2 − c 2 ) + k ( c 2 − a 2 ) k ( c 2 − a 2 ) = 1 + 1 + 1 = 3 = R.H.S. \text{L.H.S.} = \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] = \dfrac{a(ak) - b(bk)}{(a + b)((ak) - (bk))} + \dfrac{b(bk) - c(ck)}{(b + c)(bk - ck)} + \dfrac{c(ck) - a(ak)}{(c + a)(ck - ak)} \\[1em] = \dfrac{a^2k - b^2k}{k(a + b)(a - b)} + \dfrac{b^2k - c^2k}{k(b + c)(b - c)} + \dfrac{c^2k - a^2k}{k(c + a)(c - a)} \\[1em] = \dfrac{k(a^2 - b^2)}{k(a^2 - b^2)} + \dfrac{k(b^2 - c^2)}{k(b^2 - c^2)} + \dfrac{k(c^2 - a^2)}{k(c^2 - a^2)} \\[1em] = 1 + 1 + 1 \\[1em] = 3 = \text{R.H.S.} L.H.S. = ( a + b ) ( x − y ) a x − b y + ( b + c ) ( y − z ) b y − cz + ( c + a ) ( z − x ) cz − a x = ( a + b ) (( ak ) − ( bk )) a ( ak ) − b ( bk ) + ( b + c ) ( bk − c k ) b ( bk ) − c ( c k ) + ( c + a ) ( c k − ak ) c ( c k ) − a ( ak ) = k ( a + b ) ( a − b ) a 2 k − b 2 k + k ( b + c ) ( b − c ) b 2 k − c 2 k + k ( c + a ) ( c − a ) c 2 k − a 2 k = k ( a 2 − b 2 ) k ( a 2 − b 2 ) + k ( b 2 − c 2 ) k ( b 2 − c 2 ) + k ( c 2 − a 2 ) k ( c 2 − a 2 ) = 1 + 1 + 1 = 3 = R.H.S.
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) ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) = ( a b + c d + e f ) 2 (ii) ( a 3 + c 3 ) 2 ( b 3 + d 3 ) 2 = e 6 f 6 (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 (iv) b d f ( a + b b + c + d d + e + f f ) 3 = 27 ( a + b ) ( c + d ) ( e + f ) \begin{array}{ll} \text{(i)} & (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\ & = (ab + cd + ef)^2 \\ \text{(ii)} & \dfrac{(a^3 + c^3)^2}{(b^3 + d^3)^2} = \dfrac{e^6}{f^6} \\ \text{(iii)} & \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \\ & = \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\ \text{(iv)} & bdf\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^3 \\ & = 27(a + b)(c + d)(e + f) \end{array} (i) (ii) (iii) (iv) ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) = ( ab + c d + e f ) 2 ( b 3 + d 3 ) 2 ( a 3 + c 3 ) 2 = f 6 e 6 b 2 a 2 + d 2 c 2 + f 2 e 2 = b d a c + df ce + b f a e b df ( b a + b + d c + d + f e + f ) 3 = 27 ( a + b ) ( c + d ) ( e + f )
Answer
(i) Let 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 = b k , c = d k , e = f k . \therefore a = bk, c = dk, e = fk. ∴ a = bk , c = d k , e = f k .
L.H.S. = ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) = ( b 2 + d 2 + f 2 ) ( b 2 k 2 + d 2 k 2 + f 2 k 2 ) = k 2 ( b 2 + d 2 + f 2 ) ( b 2 + d 2 + f 2 ) = k 2 ( b 2 + d 2 + f 2 ) 2 . R.H.S. = ( a b + c d + e f ) 2 = ( b k . b + d k . d + f k . f ) 2 = ( b 2 k + d 2 k + f 2 k ) 2 = k 2 ( b 2 + d 2 + f 2 ) 2 . \text{L.H.S.} = (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[0.5em] = (b^2 + d^2 + f^2)(b^2k^2 + d^2k^2 + f^2k^2) \\[0.5em] = k^2(b^2 + d^2 + f^2)(b^2 + d^2 + f^2) \\[0.5em] = k^2(b^2 + d^2 + f^2)^2. \\[1em] \text{R.H.S.} = (ab + cd + ef)^2 \\[0.5em] = (bk.b + dk.d + fk .f)^2 \\[0.5em] = (b^2k + d^2k + f^2k)^2 \\[0.5em] = k^2(b^2 + d^2 + f^2)^2. L.H.S. = ( b 2 + d 2 + f 2 ) ( a 2 + c 2 + e 2 ) = ( b 2 + d 2 + f 2 ) ( b 2 k 2 + d 2 k 2 + f 2 k 2 ) = k 2 ( b 2 + d 2 + f 2 ) ( b 2 + d 2 + f 2 ) = k 2 ( b 2 + d 2 + f 2 ) 2 . R.H.S. = ( ab + c d + e f ) 2 = ( bk . b + d k . d + f k . f ) 2 = ( b 2 k + d 2 k + f 2 k ) 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) Let 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 = b k , c = d k , e = f k . \therefore a = bk, c = dk, e = fk. ∴ a = bk , c = d k , e = f k .
L.H.S. = ( a 3 + c 3 ) 2 ( b 3 + d 3 ) 2 = ( b 3 k 3 + d 3 k 3 ) 2 ( b 3 + d 3 ) 2 = [ k 3 ( b 3 + d 3 ) ] 2 ( b 3 + d 3 ) 2 = k 6 ( b 3 + d 3 ) 2 ( b 3 + d 3 ) 2 = k 6 R.H.S. = e 6 f 6 = f 6 k 6 f 6 = k 6 \text{L.H.S.} = \dfrac{(a^3 + c^3)^2}{(b^3 + d^3)^2} \\[1em] = \dfrac{(b^3k^3 + d^3k^3)^2}{(b^3 + d^3)^2} \\[1em] = \dfrac{[k^3(b^3 + d^3)]^2}{(b^3 + d^3)^2} \\[1em] = \dfrac{k^6(b^3 + d^3)^2}{(b^3 + d^3)^2} \\[1em] = k^6 \\[1em] \text{R.H.S.} = \dfrac{e^6}{f^6} \\[1em] = \dfrac{f^6k^6}{f^6} \\[1em] = k^6 L.H.S. = ( b 3 + d 3 ) 2 ( a 3 + c 3 ) 2 = ( b 3 + d 3 ) 2 ( b 3 k 3 + d 3 k 3 ) 2 = ( b 3 + d 3 ) 2 [ k 3 ( b 3 + d 3 ) ] 2 = ( b 3 + d 3 ) 2 k 6 ( b 3 + d 3 ) 2 = k 6 R.H.S. = f 6 e 6 = f 6 f 6 k 6 = k 6
Since, L.H.S. = R.H.S.
(iii) Let 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 = b k , c = d k , e = f k . \therefore a = bk, c = dk, e = fk. ∴ a = bk , c = d k , e = f k .
L.H.S. = a 2 b 2 + c 2 d 2 + e 2 f 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 R.H.S. = a c b d + c e d f + a e b f = ( b k ) d k b d + ( d k ) f k d f + ( b k ) f k b f = k 2 + k 2 + k 2 = 3 k 2 \text{L.H.S.} = \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \\[1em] = \dfrac{b^2k^2}{b^2} + \dfrac{d^2k^2}{d^2} + \dfrac{f^2k^2}{f^2} \\[1em] = k^2 + k^2 + k^2 \\[1em] = 3k^2 \\[1em] \text{R.H.S.} = \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\[1em] = \dfrac{(bk)dk}{bd} + \dfrac{(dk)fk}{df} + \dfrac{(bk)fk}{bf} \\[1em] = k^2 + k^2 + k^2 \\[1em] = 3k^2 L.H.S. = b 2 a 2 + d 2 c 2 + f 2 e 2 = b 2 b 2 k 2 + d 2 d 2 k 2 + f 2 f 2 k 2 = k 2 + k 2 + k 2 = 3 k 2 R.H.S. = b d a c + df ce + b f a e = b d ( bk ) d k + df ( d k ) f k + b f ( bk ) f k = k 2 + k 2 + k 2 = 3 k 2
Since, L.H.S. = R.H.S.
(iv) Let 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 = b k , c = d k , e = f k . \therefore a = bk, c = dk, e = fk. ∴ a = bk , c = d k , e = f k .
L.H.S. = b d f ( a + b b + c + d d + e + f f ) 3 = b d f ( b k + b b + d k + d d + f k + f f ) 3 = b d f ( k + 1 + k + 1 + k + 1 ) 3 = b d f ( 3 k + 3 ) 3 = b d f ( 3 ) 3 ( k + 1 ) 3 = 27 b d f ( k + 1 ) 3 . R.H.S. = 27 ( a + b ) ( c + d ) ( e + f ) = 27 ( b k + b ) ( d k + d ) ( f k + f ) = 27 b ( k + 1 ) d ( k + 1 ) f ( k + 1 ) = 27 b d f ( k + 1 ) 3 . \text{L.H.S.} = bdf\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^3 \\[1em] = bdf\Big(\dfrac{bk + b}{b} + \dfrac{dk + d}{d} + \dfrac{fk + f}{f}\Big)^3 \\[1em] = bdf(k + 1 + k + 1 + k + 1)^3 \\[1em] = bdf(3k + 3)^3 \\[1em] = bdf(3)^3(k + 1)^3 \\[1em] = 27bdf(k + 1)^3. \\[1em] \text{R.H.S.} = 27(a + b)(c + d)(e + f) \\[1em] = 27(bk + b)(dk + d)(fk + f) \\[1em] = 27b(k + 1)d(k + 1)f(k + 1) \\[1em] = 27bdf(k + 1)^3. L.H.S. = b df ( b a + b + d c + d + f e + f ) 3 = b df ( b bk + b + d d k + d + f f k + f ) 3 = b df ( k + 1 + k + 1 + k + 1 ) 3 = b df ( 3 k + 3 ) 3 = b df ( 3 ) 3 ( k + 1 ) 3 = 27 b df ( k + 1 ) 3 . R.H.S. = 27 ( a + b ) ( c + d ) ( e + f ) = 27 ( bk + b ) ( d k + d ) ( f k + f ) = 27 b ( k + 1 ) d ( k + 1 ) f ( k + 1 ) = 27 b df ( k + 1 ) 3 .
Since, L.H.S. = R.H.S.
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
Let a x = b y = c z = k ∴ x = k a , y = k b , z = k c L.H.S. = x 2 y z + y 2 z x + z 2 x y = k 2 a 2 k b × k c + k 2 b 2 k c × k a + k 2 c 2 k a × k b = k 2 × b c k 2 × a 2 + k 2 × a c k 2 × b 2 + k 2 × a b k 2 × c 2 = b c a 2 + a c b 2 + a b c 2 = R.H.S. \text{Let } ax = by = cz = k \\[1em] \therefore x = \dfrac{k}{a}, y = \dfrac{k}{b}, z = \dfrac{k}{c} \\[1em] \text{L.H.S.} = \dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy} \\[1em] = \dfrac{\dfrac{k^2}{a^2}}{\dfrac{k}{b} \times \dfrac{k}{c}} + \dfrac{\dfrac{k^2}{b^2}}{\dfrac{k}{c} \times \dfrac{k}{a}} + \dfrac{\dfrac{k^2}{c^2}}{\dfrac{k}{a} \times \dfrac{k}{b}} \\[1em] = \dfrac{k^2 \times bc}{k^2 \times a^2} + \dfrac{k^2 \times ac}{k^2 \times b^2} + \dfrac{k^2 \times ab}{k^2 \times c^2} \\[1em] = \dfrac{bc}{a^2} + \dfrac{ac}{b^2} + \dfrac{ab}{c^2} = \text{R.H.S.} \\[1em] Let a x = b y = cz = k ∴ x = a k , y = b k , z = c k L.H.S. = yz x 2 + z x y 2 + x y z 2 = b k × c k a 2 k 2 + c k × a k b 2 k 2 + a k × b k c 2 k 2 = k 2 × a 2 k 2 × b c + k 2 × b 2 k 2 × a c + k 2 × c 2 k 2 × ab = a 2 b c + b 2 a c + c 2 ab = R.H.S.
Since, L.H.S. = R.H.S.
If a , b , c , d a, b, c, d a , b , c , d
(i) ( 5 a + 7 b ) ( 2 c − 3 d ) = ( 5 c + 7 d ) ( 2 a − 3 b ) (ii) ( m a + n b ) : b = ( m c + n d ) : d (iii) ( a 4 + c 4 ) : ( b 4 + d 4 ) = a 2 c 2 : b 2 d 2 (iv) a 2 + a b c 2 + c d = b 2 − 2 a b d 2 − 2 c d (v) ( a + c ) 3 ( b + d ) 3 = a ( a − c ) 2 b ( b − d ) 2 (vi) a 2 + a b + b 2 a 2 − a b + b 2 = c 2 + c d + d 2 c 2 − c d + d 2 (vii) a 2 + b 2 c 2 + d 2 = a b + a d − b c b c + c d − a d (viii) a b c d ( 1 a 2 + 1 b 2 + 1 c 2 + 1 d 2 ) = a 2 + b 2 + c 2 + d 2 . \begin{array}{ll} \text{(i)} & (5a + 7b)(2c - 3d) \\ & = (5c + 7d)(2a - 3b) \\ \text{(ii)} & (ma + nb) : b \\ & = (mc + nd) : d \\ \text{(iii)} & (a^4 + c^4) : (b^4 + d^4) \\ & = a^2c^2 : b^2d^2 \\ \text{(iv)} & \dfrac{a^2 + ab}{c^2 + cd} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \text{(v)} & \dfrac{(a + c)^3}{(b + d)^3} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \text{(vi)} & \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\ & = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\ \text{(vii)} & \dfrac{a^2 + b^2}{c^2 + d^2} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] \text{(viii)} & abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \\ & = a^2 + b^2 + c^2 + d^2. \end{array} (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) ( 5 a + 7 b ) ( 2 c − 3 d ) = ( 5 c + 7 d ) ( 2 a − 3 b ) ( ma + nb ) : b = ( m c + n d ) : d ( a 4 + c 4 ) : ( b 4 + d 4 ) = a 2 c 2 : b 2 d 2 c 2 + c d a 2 + ab = d 2 − 2 c d b 2 − 2 ab ( b + d ) 3 ( a + c ) 3 = b ( b − d ) 2 a ( a − c ) 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 + d 2 a 2 + b 2 = b c + c d − a d ab + a d − b c ab c d ( a 2 1 + b 2 1 + c 2 1 + d 2 1 ) = a 2 + b 2 + c 2 + d 2 .
Answer
(i) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = ( 5 a + 7 b ) ( 2 c − 3 d ) = ( 5 b k + 7 b ) ( 2 d k − 3 d ) = b d ( 5 k + 7 ) ( 2 k − 3 ) R.H.S. = ( 5 c + 7 d ) ( 2 a − 3 b ) = ( 5 d k + 7 d ) ( 2 b k − 3 b ) = b d ( 5 k + 7 ) ( 2 k − 3 ) . \text{L.H.S.} = (5a + 7b)(2c - 3d) \\[0.5em] = (5bk + 7b)(2dk - 3d) \\[0.5em] = bd(5k + 7)(2k - 3) \\[1em] \text{R.H.S.} = (5c + 7d)(2a - 3b) \\[0.5em] = (5dk + 7d)(2bk - 3b) \\[0.5em] = bd(5k + 7)(2k - 3). L.H.S. = ( 5 a + 7 b ) ( 2 c − 3 d ) = ( 5 bk + 7 b ) ( 2 d k − 3 d ) = b d ( 5 k + 7 ) ( 2 k − 3 ) R.H.S. = ( 5 c + 7 d ) ( 2 a − 3 b ) = ( 5 d k + 7 d ) ( 2 bk − 3 b ) = b d ( 5 k + 7 ) ( 2 k − 3 ) .
Since, L.H.S. = R.H.S.
(ii) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = ( m a + n b ) : b = m a + n b b \text{L.H.S.} = (ma + nb) : b \\[0.5em] = \dfrac{ma + nb}{b} \\[0.5em] L.H.S. = ( ma + nb ) : b = b ma + nb
Putting value of a = bk,
= m b k + n b b = m k + n . R.H.S. = ( m c + n d ) : d = \dfrac{mbk + nb}{b} \\[0.5em] = mk + n. \\[1em] \text{R.H.S.} = (mc + nd) : d \\[0.5em] = b mbk + nb = mk + n . R.H.S. = ( m c + n d ) : d
Putting value of c = dk,
= m d k + n d d = m k + n . = \dfrac{mdk + nd}{d} \\[0.5em] = mk + n. = d m d k + n d = mk + n .
Since, L.H.S. = R.H.S.
(iii) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = ( a 4 + c 4 ) : ( b 4 + d 4 ) = a 4 + c 4 b 4 + d 4 = k 4 b 4 + k 4 d 4 b 4 + d 4 = k 4 . R.H.S. = a 2 c 2 : b 2 d 2 = a 2 c 2 b 2 d 2 = ( b k ) 2 ( d k ) 2 b 2 d 2 = k 4 b 2 d 2 b 2 d 2 = k 4 . \text{L.H.S.} = (a^4 + c^4) : (b^4 + d^4) \\[1em] = \dfrac{a^4 + c^4}{b^4 + d^4} \\[1em] = \dfrac{k^4b^4 + k^4d^4}{b^4 + d^4} \\[1em] = k^4. \\[1em] \text{R.H.S.} = a^2c^2 : b^2d^2 \\[1em] = \dfrac{a^2c^2}{b^2d^2} \\[1em] = \dfrac{(bk)^2(dk)^2}{b^2d^2} \\[1em] = \dfrac{k^4b^2d^2}{b^2d^2} \\[1em] = k^4. L.H.S. = ( a 4 + c 4 ) : ( b 4 + d 4 ) = b 4 + d 4 a 4 + c 4 = b 4 + d 4 k 4 b 4 + k 4 d 4 = k 4 . R.H.S. = a 2 c 2 : b 2 d 2 = b 2 d 2 a 2 c 2 = b 2 d 2 ( bk ) 2 ( d k ) 2 = b 2 d 2 k 4 b 2 d 2 = k 4 .
Since, L.H.S. = R.H.S.
(iv) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = a 2 + a b c 2 + c d = b 2 k 2 + b 2 k d 2 k 2 + d 2 k = b 2 k ( k + 1 ) d 2 k ( k + 1 ) = b 2 d 2 R.H.S. = b 2 − 2 a b d 2 − 2 c d = b 2 − 2 b 2 k d 2 − 2 d 2 k = b 2 ( 1 − 2 k ) d 2 ( 1 − 2 k ) = b 2 d 2 . \text{L.H.S.} = \dfrac{a^2 + ab}{c^2 + cd} \\[1em] = \dfrac{b^2k^2 + b^2k}{d^2k^2 + d^2k} \\[1em] = \dfrac{b^2k(k + 1)}{d^2k(k + 1)} \\[1em] = \dfrac{b^2}{d^2} \\[1em] \text{R.H.S.} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] = \dfrac{b^2 - 2b^2k}{d^2 - 2d^2k} \\[1em] = \dfrac{b^2(1 - 2k)}{d^2(1 - 2k)} \\[1em] = \dfrac{b^2}{d^2}. L.H.S. = c 2 + c d a 2 + ab = d 2 k 2 + d 2 k b 2 k 2 + b 2 k = d 2 k ( k + 1 ) b 2 k ( k + 1 ) = d 2 b 2 R.H.S. = d 2 − 2 c d b 2 − 2 ab = d 2 − 2 d 2 k b 2 − 2 b 2 k = d 2 ( 1 − 2 k ) b 2 ( 1 − 2 k ) = d 2 b 2 .
Since, L.H.S. = R.H.S.
(v) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = ( a + c ) 3 ( b + d ) 3 = ( b k + d k ) 3 ( b + d ) 3 = k 3 ( b + d ) 3 ( b + d ) 3 = k 3 R.H.S. = a ( a − c ) 2 b ( b − d ) 2 = b k ( b k − d k ) 2 b ( b − d ) 2 = b k ( k 2 ( b − d ) 2 ) b ( b − d ) 2 = b k 3 ( b − d ) 2 b ( b − d ) 2 = k 3 . \text{L.H.S.} = \dfrac{(a + c)^3}{(b + d)^3} \\[1em] = \dfrac{(bk + dk)^3}{(b + d)^3} \\[1em] = \dfrac{k^3(b + d)^3}{(b + d)^3} \\[1em] = k^3 \\[1em] \text{R.H.S.} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] = \dfrac{bk(bk - dk)^2}{b(b - d)^2} \\[1em] = \dfrac{bk(k^2(b - d)^2)}{b(b - d)^2} \\[1em] = \dfrac{bk^3(b - d)^2}{b(b - d)^2} \\[1em] = k^3. L.H.S. = ( b + d ) 3 ( a + c ) 3 = ( b + d ) 3 ( bk + d k ) 3 = ( b + d ) 3 k 3 ( b + d ) 3 = k 3 R.H.S. = b ( b − d ) 2 a ( a − c ) 2 = b ( b − d ) 2 bk ( bk − d k ) 2 = b ( b − d ) 2 bk ( k 2 ( b − d ) 2 ) = b ( b − d ) 2 b k 3 ( b − d ) 2 = k 3 .
Since, L.H.S. = R.H.S.
(vi) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = a 2 + a b + b 2 a 2 − a b + b 2 = b 2 k 2 + ( b k ) b + b 2 b 2 k 2 − ( b k ) b + b 2 = b 2 k 2 + b 2 k + b 2 b 2 k 2 − b 2 k + b 2 = b 2 ( k 2 + k + 1 ) b 2 ( k 2 − k + 1 ) = k 2 + k + 1 k 2 − k + 1 R.H.S. = c 2 + c d + d 2 c 2 − c d + d 2 = d 2 k 2 + ( d k ) d + d 2 d 2 k 2 − ( d k ) d + d 2 = d 2 ( k 2 + k + 1 ) d 2 ( k 2 − k + 1 ) = k 2 + k + 1 k 2 − k + 1 . \text{L.H.S.} = \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] = \dfrac{b^2k^2 + (bk)b + b^2}{b^2k^2 - (bk)b + b^2} \\[1em] = \dfrac{b^2k^2 + b^2k + b^2}{b^2k^2 - b^2k + b^2} \\[1em] = \dfrac{b^2(k^2 + k + 1)}{b^2(k^2 - k + 1)} \\[1em] = \dfrac{k^2 + k + 1}{k^2 - k + 1} \\[1em] \text{R.H.S.} = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\[1em] = \dfrac{d^2k^2 + (dk)d + d^2}{d^2k^2 - (dk)d + d^2} \\[1em] = \dfrac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] = \dfrac{k^2 + k + 1}{k^2 - k + 1}. L.H.S. = a 2 − ab + b 2 a 2 + ab + b 2 = b 2 k 2 − ( bk ) b + b 2 b 2 k 2 + ( bk ) b + b 2 = b 2 k 2 − b 2 k + b 2 b 2 k 2 + b 2 k + b 2 = b 2 ( k 2 − k + 1 ) b 2 ( k 2 + k + 1 ) = k 2 − k + 1 k 2 + k + 1 R.H.S. = c 2 − c d + d 2 c 2 + c d + d 2 = d 2 k 2 − ( d k ) d + d 2 d 2 k 2 + ( d k ) d + 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.
(vii) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = a 2 + b 2 c 2 + d 2 = b 2 k 2 + b 2 d 2 k 2 + d 2 = b 2 ( k 2 + 1 ) d 2 ( k 2 + 1 ) = b 2 d 2 R.H.S. = a b + a d − b c b c + c d − a d = ( b k ) b + ( b k ) d − b ( d k ) b ( d k ) + ( d k ) d − ( b k ) d = b 2 k + b d k − b d k b d k + d 2 k − b d k = b 2 k d 2 k = b 2 d 2 . \text{L.H.S.} = \dfrac{a^2 + b^2}{c^2 + d^2} \\[1em] = \dfrac{b^2k^2 + b^2}{d^2k^2 + d^2} \\[1em] = \dfrac{b^2(k^2 + 1)}{d^2(k^2 + 1)} \\[1em] = \dfrac{b^2}{d^2} \\[1em] \text{R.H.S.} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] = \dfrac{(bk)b + (bk)d - b(dk)}{b(dk) + (dk)d - (bk)d} \\[1em] = \dfrac{b^2k + bdk - bdk}{bdk + d^2k - bdk} \\[1em] = \dfrac{b^2k}{d^2k} \\[1em] = \dfrac{b^2}{d^2}. L.H.S. = c 2 + d 2 a 2 + b 2 = d 2 k 2 + d 2 b 2 k 2 + b 2 = d 2 ( k 2 + 1 ) b 2 ( k 2 + 1 ) = d 2 b 2 R.H.S. = b c + c d − a d ab + a d − b c = b ( d k ) + ( d k ) d − ( bk ) d ( bk ) b + ( bk ) d − b ( d k ) = b d k + d 2 k − b d k b 2 k + b d k − b d k = d 2 k b 2 k = d 2 b 2 .
Since, L.H.S. = R.H.S.
(viii) a, b, c, d are in proportion
∴ a b = c d = k \therefore \dfrac{a}{b} = \dfrac{c}{d} = k ∴ b a = d c = k
Hence, a = bk, c = dk.
L.H.S. = a b c d ( 1 a 2 + 1 b 2 + 1 c 2 + 1 d 2 ) = ( b k ) b ( d k ) d ( 1 b 2 k 2 + 1 b 2 + 1 d 2 k 2 + 1 d 2 ) = b 2 d 2 k 2 ( d 2 + d 2 k 2 + b 2 + b 2 k 2 b 2 d 2 k 2 ) = d 2 ( 1 + k 2 ) + b 2 ( 1 + k 2 ) = ( d 2 + b 2 ) ( 1 + k 2 ) . R.H.S. = a 2 + b 2 + c 2 + d 2 = b 2 k 2 + b 2 + d 2 k 2 + d 2 = b 2 ( k 2 + 1 ) + d 2 ( k 2 + 1 ) = ( d 2 + b 2 ) ( 1 + k 2 ) . \text{L.H.S.} = abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \\[1em] = (bk)b(dk)d\Big(\dfrac{1}{b^2k^2} + \dfrac{1}{b^2} + \dfrac{1}{d^2k^2} + \dfrac{1}{d^2}\Big) \\[1em] = b^2d^2k^2\Big(\dfrac{d^2 + d^2k^2 + b^2 + b^2k^2}{b^2d^2k^2}\Big) \\[1em] = d^2(1 + k^2) + b^2(1 + k^2) \\[1em] = (d^2 + b^2)(1 + k^2). \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 + d^2 \\[1em] = b^2k^2 + b^2 + d^2k^2 + d^2 \\[1em] = b^2(k^2 + 1) + d^2(k^2 + 1) \\[1em] = (d^2 + b^2)(1 + k^2). L.H.S. = ab c d ( a 2 1 + b 2 1 + c 2 1 + d 2 1 ) = ( bk ) b ( d k ) d ( b 2 k 2 1 + b 2 1 + d 2 k 2 1 + d 2 1 ) = b 2 d 2 k 2 ( b 2 d 2 k 2 d 2 + d 2 k 2 + b 2 + b 2 k 2 ) = d 2 ( 1 + k 2 ) + b 2 ( 1 + k 2 ) = ( d 2 + b 2 ) ( 1 + k 2 ) . R.H.S. = a 2 + b 2 + c 2 + d 2 = b 2 k 2 + b 2 + d 2 k 2 + d 2 = b 2 ( k 2 + 1 ) + d 2 ( k 2 + 1 ) = ( d 2 + b 2 ) ( 1 + k 2 ) .
Since, L.H.S. = R.H.S.
If x, y, z are in continued proportion, prove that : ( x + y ) 2 ( y + z ) 2 = x z . \dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z}. ( y + z ) 2 ( x + y ) 2 = z x .
Answer
Since, x, y, z are in continued proportion
∴ x y = y z = k ⇒ y = z k and x = y k = z k 2 L.H.S. = ( x + y ) 2 ( y + z ) 2 = ( z k 2 + z k ) 2 ( z k + z ) 2 = z 2 k 4 + z 2 k 2 + 2 z 2 k 3 z 2 k 2 + z 2 + 2 z 2 k = z 2 k 2 ( k 2 + 1 + 2 k ) z 2 ( k 2 + 1 + 2 k ) = k 2 . R.H.S. = x z = z k 2 z = k 2 . \therefore \dfrac{x}{y} = \dfrac{y}{z} = k \\[1em] \Rightarrow y = zk \text{ and } x = yk = zk^2 \\[1em] \text{L.H.S.} = \dfrac{(x + y)^2}{(y + z)^2} \\[1em] = \dfrac{(zk^2 + zk)^2}{(zk + z)^2} \\[1em] = \dfrac{z^2k^4 + z^2k^2 + 2z^2k^3}{z^2k^2 + z^2 + 2z^2k} \\[1em] = \dfrac{z^2k^2(k^2 + 1 + 2k)}{z^2(k^2 + 1 + 2k)} \\[1em] = k^2. \\[1em] \text{R.H.S.} = \dfrac{x}{z} \\[1em] = \dfrac{zk^2}{z} = k^2. ∴ y x = z y = k ⇒ y = z k and x = y k = z k 2 L.H.S. = ( y + z ) 2 ( x + y ) 2 = ( z k + z ) 2 ( z k 2 + z k ) 2 = z 2 k 2 + z 2 + 2 z 2 k z 2 k 4 + z 2 k 2 + 2 z 2 k 3 = z 2 ( k 2 + 1 + 2 k ) z 2 k 2 ( k 2 + 1 + 2 k ) = k 2 . R.H.S. = z x = z z k 2 = k 2 .
Since, L.H.S. = R.H.S.
If a, b, c are in continued proportion, prove that :
p a 2 + q a b + r b 2 p b 2 + q b c + r c 2 = a c . \dfrac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \dfrac{a}{c}. p b 2 + q b c + r c 2 p a 2 + q ab + r b 2 = c a .
Answer
Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = p a 2 + q a b + r b 2 p b 2 + q b c + r c 2 = p ( c k 2 ) 2 + q ( c k 2 ) ( c k ) + r ( c k ) 2 p ( c k ) 2 + q ( c k ) c + r c 2 = p c 2 k 4 + q c 2 k 3 + r c 2 k 2 p c 2 k 2 + q c 2 k + r c 2 = c 2 k 2 ( p k 2 + q k + r ) c 2 ( p k 2 + q k + r ) = k 2 R.H.S. = a c = c k 2 c = k 2 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \dfrac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} \\[1em] = \dfrac{p(ck^2)^2 + q(ck^2)(ck) + r(ck)^2}{p(ck)^2 + q(ck)c + rc^2} \\[1em] = \dfrac{pc^2k^4 + qc^2k^3 + rc^2k^2}{pc^2k^2 + qc^2k + rc^2} \\[1em] = \dfrac{c^2k^2(pk^2 + qk + r)}{c^2(pk^2 + qk + r)} \\[1em] = k^2 \\[1em] \text{R.H.S.} = \dfrac{a}{c} \\[1em] = \dfrac{ck^2}{c} \\[1em] = k^2. ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = p b 2 + q b c + r c 2 p a 2 + q ab + r b 2 = p ( c k ) 2 + q ( c k ) c + r c 2 p ( c k 2 ) 2 + q ( c k 2 ) ( c k ) + r ( c k ) 2 = p c 2 k 2 + q c 2 k + r c 2 p c 2 k 4 + q c 2 k 3 + r c 2 k 2 = c 2 ( p k 2 + q k + r ) c 2 k 2 ( p k 2 + q k + r ) = k 2 R.H.S. = c a = c c k 2 = k 2 .
Since, L.H.S. = R.H.S. hence proved that,
p a 2 + q a b + r b 2 p b 2 + q b c + r c 2 = a c . \dfrac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \dfrac{a}{c}. p b 2 + q b c + r c 2 p a 2 + q ab + r b 2 = c a .
If a , b , c a, b, c a , b , c
(i) a + b b + c = a 2 ( b − c ) b 2 ( a − b ) (ii) 1 a 3 + 1 b 3 + 1 c 3 = a b 2 c 2 + b c 2 a 2 + c a 2 b 2 (iii) a : c = ( a 2 + b 2 ) : ( b 2 + c 2 ) (iv) a 2 b 2 c 2 ( a − 4 + b − 4 + c − 4 ) = b − 2 ( a 4 + b 4 + c 4 ) (v) a b c ( a + b + c ) 3 = ( a b + b c + c a ) 3 (vi) ( a + b + c ) ( a − b + c ) = a 2 + b 2 + c 2 . \begin{array}{ll} \text{(i)} & \dfrac{a + b}{b + c} = \dfrac{a^2(b - c)}{b^2(a - b)} \\ \text{(ii)} & \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \\ & = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} \\ \text{(iii)} & a : c = (a^2 + b^2) : (b^2 + c^2) \\ \text{(iv)} & a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \\ & = b^{-2}(a^4 + b^4 + c^4) \\ \text{(v)} & abc(a + b + c)^3 \\ & = (ab + bc + ca)^3 \\ \text{(vi)} & (a + b + c)(a - b + c) \\ & = a^2 + b^2 + c^2. \end{array} (i) (ii) (iii) (iv) (v) (vi) b + c a + b = b 2 ( a − b ) a 2 ( b − c ) a 3 1 + b 3 1 + c 3 1 = b 2 c 2 a + c 2 a 2 b + a 2 b 2 c a : c = ( a 2 + b 2 ) : ( b 2 + c 2 ) a 2 b 2 c 2 ( a − 4 + b − 4 + c − 4 ) = b − 2 ( a 4 + b 4 + c 4 ) ab c ( a + b + c ) 3 = ( ab + b c + c a ) 3 ( a + b + c ) ( a − b + c ) = a 2 + b 2 + c 2 .
Answer
(i) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = a + b b + c = c k 2 + c k c k + c = c k ( k + 1 ) c ( k + 1 ) = k . R.H.S. = 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 − 1 ) c 2 k 2 c k ( k − 1 ) = c 3 k 4 ( k − 1 ) c 3 k 3 ( k − 1 ) = k . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \dfrac{a + b}{b + c} \\[1em] = \dfrac{ck^2 + ck}{ck + c} \\[1em] = \dfrac{ck(k + 1)}{c(k + 1)} \\[1em] = k. \\[1em] \text{R.H.S.} = \dfrac{a^2(b - c)}{b^2(a - b)} \\[1em] = \dfrac{(ck^2)^2(ck - c)}{(ck)^2(ck^2 - ck)} \\[1em] = \dfrac{c^2k^4c(k - 1)}{c^2k^2ck(k - 1)} \\[1em] = \dfrac{c^3k^4(k - 1)}{c^3k^3(k - 1)} \\[1em] = k. ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = b + c a + b = c k + c c k 2 + c k = c ( k + 1 ) c k ( k + 1 ) = k . R.H.S. = 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 ( k − 1 ) c 2 k 4 c ( k − 1 ) = 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) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = 1 a 3 + 1 b 3 + 1 c 3 = 1 ( c k 2 ) 3 + 1 ( c k ) 3 + 1 c 3 = 1 c 3 k 6 + 1 c 3 k 3 + 1 c 3 = 1 c 3 ( 1 k 6 + 1 k 3 + 1 ) R.H.S. = a b 2 c 2 + b c 2 a 2 + c a 2 b 2 = c k 2 ( c k ) 2 c 2 + c k c 2 ( c k 2 ) 2 + c ( c k 2 ) 2 ( c k ) 2 = c k 2 c 4 k 2 + c k c 4 k 4 + c c 4 k 6 = 1 c 3 + 1 c 3 k 3 + 1 c 3 k 6 = 1 c 3 ( 1 + 1 k 3 + 1 k 6 ) \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{(ck^2)^3} + \dfrac{1}{(ck)^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{c^3k^6} + \dfrac{1}{c^3k^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{c^3}\Big(\dfrac{1}{k^6} + \dfrac{1}{k^3} + 1\Big) \\[1em] \text{R.H.S.} = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} \\[1em] = \dfrac{ck^2}{(ck)^2c^2} + \dfrac{ck}{c^2(ck^2)^2} + \dfrac{c}{(ck^2)^2(ck)^2} \\[1em] = \dfrac{ck^2}{c^4k^2} + \dfrac{ck}{c^4k^4} + \dfrac{c}{c^4k^6} \\[1em] = \dfrac{1}{c^3} + \dfrac{1}{c^3k^3} + \dfrac{1}{c^3k^6} \\[1em] = \dfrac{1}{c^3}\Big(1 + \dfrac{1}{k^3} + \dfrac{1}{k^6}\Big) \\[1em] ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = a 3 1 + b 3 1 + c 3 1 = ( c k 2 ) 3 1 + ( c k ) 3 1 + c 3 1 = c 3 k 6 1 + c 3 k 3 1 + c 3 1 = c 3 1 ( k 6 1 + k 3 1 + 1 ) R.H.S. = b 2 c 2 a + c 2 a 2 b + a 2 b 2 c = ( c k ) 2 c 2 c k 2 + c 2 ( c k 2 ) 2 c k + ( c k 2 ) 2 ( c k ) 2 c = c 4 k 2 c k 2 + c 4 k 4 c k + c 4 k 6 c = c 3 1 + c 3 k 3 1 + c 3 k 6 1 = c 3 1 ( 1 + k 3 1 + k 6 1 )
Since, L.H.S. = R.H.S. hence proved that,
1 a 3 + 1 b 3 + 1 c 3 = a b 2 c 2 + b c 2 a 2 + c a 2 b 2 \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} a 3 1 + b 3 1 + c 3 1 = b 2 c 2 a + c 2 a 2 b + a 2 b 2 c
(iii) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = a : c = a c = c k 2 c = k 2 . R.H.S. = ( a 2 + b 2 ) : ( b 2 + c 2 ) = a 2 + b 2 b 2 + c 2 = ( ( c k 2 ) 2 + ( c k ) 2 ) ( ( c k ) 2 + c 2 ) = c 2 k 4 + c 2 k 2 c 2 k 2 + c 2 = c 2 k 2 ( k 2 + 1 ) c 2 ( k 2 + 1 ) = k 2 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a : c \\[1em] = \dfrac{a}{c} \\[1em] = \dfrac{ck^2}{c} \\[1em] = k^2. \\[1em] \text{R.H.S.} = (a^2 + b^2) : (b^2 + c^2) \\[1em] = \dfrac{a^2 + b^2}{b^2 + c^2} \\[1em] = \dfrac{((ck^2)^2 + (ck)^2)}{((ck)^2 + c^2)} \\[1em] = \dfrac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} \\[1em] = \dfrac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} \\[1em] = k^2. \\[1em] ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = a : c = c a = c c k 2 = k 2 . R.H.S. = ( a 2 + b 2 ) : ( b 2 + c 2 ) = b 2 + c 2 a 2 + b 2 = (( c k ) 2 + c 2 ) (( c k 2 ) 2 + ( c k ) 2 ) = c 2 k 2 + c 2 c 2 k 4 + c 2 k 2 = c 2 ( k 2 + 1 ) c 2 k 2 ( k 2 + 1 ) = k 2 .
Since, L.H.S = R.H.S hence proved that,
a : c = (a2 + b2 ) : (b2 + c2 ).
(iv) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = a 2 b 2 c 2 ( a − 4 + b − 4 + c − 4 ) = ( c k 2 ) 2 ( c k ) 2 c 2 ( ( c k 2 ) − 4 + ( c k ) − 4 + c − 4 ) = c 2 k 4 c 2 k 2 c 2 ( c − 4 k − 8 + c − 4 k − 4 + c − 4 ) = c 6 k 6 c − 4 ( k − 8 + k − 4 + 1 ) = c 2 k 6 ( 1 k 8 + 1 k 4 + 1 ) = c 2 k 6 ( 1 + k 4 + k 8 k 8 ) = c 2 k 6 k 8 ( 1 + k 4 + k 8 ) = c 2 k 2 ( 1 + k 4 + k 8 ) R.H.S. = b − 2 ( a 4 + b 4 + c 4 ) = ( c k ) − 2 ( ( c k 2 ) 4 + ( c k ) 4 + c 4 ) = c − 2 k − 2 ( c 4 k 8 + c 4 k 4 + c 4 ) = c − 2 k − 2 c 4 ( k 8 + k 4 + 1 ) = c 2 k − 2 ( k 8 + k 4 + 1 ) = c 2 k 2 ( 1 + k 4 + k 8 ) . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \\[1em] = (ck^2)^2(ck)^2c^2((ck^2)^{-4} + (ck)^{-4} + c^{-4}) \\[1em] = c^2k^4c^2k^2c^2(c^{-4}k^{-8} + c^{-4}k^{-4} + c^{-4}) \\[1em] = c^6k^6c^{-4}(k^{-8} + k^{-4} + 1) \\[1em] = c^2k^6\big(\dfrac{1}{k^8} + \dfrac{1}{k^4} + 1\big) \\[1em] = c^2k^6\big(\dfrac{1 + k^4 + k^8}{k^8}\big) \\[1em] = \dfrac{c^2k^6}{k^8}(1 + k^4 + k^8) \\[1em] = \dfrac{c^2}{k^2}(1 + k^4 + k^8) \\[1em] \text{R.H.S.} = b^{-2}(a^4 + b^4 + c^4) \\[1em] = (ck)^{-2}((ck^2)^4 + (ck)^4 + c^4) \\[1em] = c^{-2}k^{-2}(c^4k^8 + c^4k^4 + c^4) \\[1em] = c^{-2}k^{-2}c^4(k^8 + k^4 + 1) \\[1em] = c^2k^{-2}(k^8 + k^4 + 1) \\[1em] = \dfrac{c^2}{k^2}(1 + k^4 + k^8). \\[1em] ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = a 2 b 2 c 2 ( a − 4 + b − 4 + c − 4 ) = ( c k 2 ) 2 ( c k ) 2 c 2 (( c k 2 ) − 4 + ( c k ) − 4 + c − 4 ) = c 2 k 4 c 2 k 2 c 2 ( c − 4 k − 8 + c − 4 k − 4 + c − 4 ) = c 6 k 6 c − 4 ( k − 8 + k − 4 + 1 ) = c 2 k 6 ( k 8 1 + k 4 1 + 1 ) = c 2 k 6 ( k 8 1 + k 4 + k 8 ) = k 8 c 2 k 6 ( 1 + k 4 + k 8 ) = k 2 c 2 ( 1 + k 4 + k 8 ) R.H.S. = b − 2 ( a 4 + b 4 + c 4 ) = ( c k ) − 2 (( c k 2 ) 4 + ( c k ) 4 + c 4 ) = c − 2 k − 2 ( c 4 k 8 + c 4 k 4 + c 4 ) = c − 2 k − 2 c 4 ( k 8 + k 4 + 1 ) = c 2 k − 2 ( k 8 + k 4 + 1 ) = k 2 c 2 ( 1 + k 4 + k 8 ) .
Since, L.H.S. = R.H.S. hence proved that,
a2 b2 c2 (a-4 + b-4 + c-4 ) = b-2 (a4 + b4 + c4 ).
(v) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = a b c ( a + b + c ) 3 = ( c k 2 ) ( c k ) c ( c k 2 + c k + c ) 3 = c 3 k 3 ( c 3 ( k 2 + k + 1 ) 3 ) = c 6 k 3 ( k 2 + k + 1 ) 3 R.H.S. = ( a b + b c + c a ) 3 = ( ( c k 2 ) ( c k ) + ( c k ) ( c ) + ( c ) ( c k 2 ) ) 3 = ( c 2 k 3 + c 2 k + c 2 k 2 ) 3 = ( ( c 2 k ) 3 ( k 2 + 1 + k ) 3 ) = c 6 k 3 ( k 2 + k + 1 ) 3 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = abc(a + b + c)^3 \\[1em] = (ck^2)(ck)c(ck^2 + ck + c)^3 \\[1em] = c^3k^3(c^3(k^2 + k + 1)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3 \\[1em] \text{R.H.S.} = (ab + bc + ca)^3 \\[1em] = ((ck^2)(ck) + (ck)(c) + (c)(ck^2))^3 \\[1em] = (c^2k^3 + c^2k + c^2k^2)^3 \\[1em] = ((c^2k)^3(k^2 + 1 + k)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3. \\[1em] ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = ab c ( a + b + c ) 3 = ( c k 2 ) ( c k ) c ( c k 2 + c k + c ) 3 = c 3 k 3 ( c 3 ( k 2 + k + 1 ) 3 ) = c 6 k 3 ( k 2 + k + 1 ) 3 R.H.S. = ( ab + b c + c a ) 3 = (( c k 2 ) ( c k ) + ( c k ) ( c ) + ( c ) ( c k 2 ) ) 3 = ( c 2 k 3 + c 2 k + c 2 k 2 ) 3 = (( c 2 k ) 3 ( k 2 + 1 + k ) 3 ) = c 6 k 3 ( k 2 + k + 1 ) 3 .
Since L.H.S. = R.H.S. hence proved that,
abc(a + b + c)3 = (ab + bc + ca)3 .
(vi) Since, a, b, c are in continued proportion
∴ a b = b c = k ⇒ b = c k and a = b k = c k 2 L.H.S. = ( 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 ) = c 2 ( k 4 − k 3 + k 2 + k 3 − k 2 + k + k 2 − k + 1 ) = c 2 ( k 4 − k 3 + k 2 + k 3 − k 2 + k + k 2 − k + 1 ) = c 2 ( k 4 + k 2 + 1 ) R.H.S. = a 2 + b 2 + c 2 = ( c k 2 ) 2 + ( c k ) 2 + c 2 = c 2 k 4 + c 2 k 2 + c 2 = c 2 ( k 4 + k 2 + 1 ) . \therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = (a + b + c)(a - b + c) \\[1em] = (ck^2 + ck + c)(ck^2 - ck + c) \\[1em] = c(k^2 + k + 1)c(k^2 - k + 1) \\[1em] = c^2(k^4 - k^3 + k^2 + k^3 - k^2 + k + k^2 - k + 1) \\[1em] = c^2(k^4 - \bcancel{k^3} + \bcancel{k^2} + \bcancel{k^3} - \bcancel{k^2} + \bcancel{k} + k^2 - \bcancel{k} + 1) \\[1em] = c^2(k^4 + k^2 + 1) \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 \\[1em] = (ck^2)^2 + (ck)^2 + c^2 \\[1em] = c^2k^4 + c^2k^2 + c^2 \\[1em] = c^2(k^4 + k^2 + 1). \\[1em] ∴ b a = c b = k ⇒ b = c k and a = bk = c k 2 L.H.S. = ( 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 ) = c 2 ( k 4 − k 3 + k 2 + k 3 − k 2 + k + k 2 − k + 1 ) = c 2 ( k 4 − k 3 + k 2 + k 3 − k 2 + k + k 2 − k + 1 ) = c 2 ( k 4 + k 2 + 1 ) R.H.S. = a 2 + b 2 + c 2 = ( c k 2 ) 2 + ( c k ) 2 + c 2 = c 2 k 4 + c 2 k 2 + c 2 = c 2 ( k 4 + k 2 + 1 ) .
Since, L.H.S. = R.H.S. hence proved that,
(a + b + c)(a - b + c) = a2 + b2 + c2 .
If a, b, c, d are in continued proportion, prove that :
(i) a 3 + b 3 + c 3 b 3 + c 3 + d 3 = a d \dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d} b 3 + c 3 + d 3 a 3 + b 3 + c 3 = d a
(ii) (a2 - b2 )(c2 - d2 ) = (b2 - c2 )2
(iii) (a + d)(b + c) - (a + c)(b + d) = (b - c)2
(iv) a : d = triplicate ratio of (a - b) : (b - c)
(v) ( a − b c + a − c b ) 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}{b}\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 + b a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 − b 2 1 ) .
Answer
(i) Since, a, b, c, d are in continued proportion
∴ a b = b c = c d = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = b k = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = a 3 + b 3 + c 3 b 3 + c 3 + d 3 = ( d k 3 ) 3 + ( d k 2 ) 3 + ( d k ) 3 ( d k 2 ) 3 + ( d k ) 3 + d 3 = d 3 k 9 + d 3 k 6 + d 3 k 3 d 3 k 6 + d 3 k 3 + d 3 = d 3 k 3 ( k 6 + k 3 + 1 ) d 3 ( k 6 + k 3 + 1 ) = k 3 . R.H.S. = a d = d k 3 d = k 3 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} \\[1em] = \dfrac{(dk^3)^3 + (dk^2)^3 + (dk)^3}{(dk^2)^3 + (dk)^3 + d^3} \\[1em] = \dfrac{d^3k^9 + d^3k^6 + d^3k^3}{d^3k^6 + d^3k^3 + d^3} \\[1em] = \dfrac{d^3k^3(k^6 + k^3 + 1)}{d^3(k^6 + k^3 + 1)} \\[1em] = k^3. \\[1em] \text{R.H.S.} = \dfrac{a}{d} \\[1em] = \dfrac{dk^3}{d} \\[1em] = k^3. \\[1em] ∴ b a = c b = d c = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = bk = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = b 3 + c 3 + d 3 a 3 + b 3 + c 3 = ( d k 2 ) 3 + ( d k ) 3 + d 3 ( d k 3 ) 3 + ( d k 2 ) 3 + ( d k ) 3 = d 3 k 6 + d 3 k 3 + d 3 d 3 k 9 + d 3 k 6 + d 3 k 3 = d 3 ( k 6 + k 3 + 1 ) d 3 k 3 ( k 6 + k 3 + 1 ) = k 3 . R.H.S. = d a = d d k 3 = k 3 .
Since, L.H.S. = R.H.S. hence, proved that,
a 3 + b 3 + c 3 b 3 + c 3 + d 3 = a d \dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d} b 3 + c 3 + d 3 a 3 + b 3 + c 3 = d a
(ii) Since, a, b, c, d are in continued proportion
∴ a b = b c = c d = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = b k = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( a 2 − b 2 ) ( c 2 − d 2 ) = [ ( d k 3 ) 2 − ( d k 2 ) 2 ] [ ( d k ) 2 − ( d 2 ) ] = ( d 2 k 6 − d 2 k 4 ) ( d 2 k 2 − d 2 ) = d 2 k 4 ( k 2 − 1 ) d 2 ( k 2 − 1 ) = d 4 k 4 ( k 2 − 1 ) 2 . R.H.S. = ( b 2 − c 2 ) 2 = ( ( d k 2 ) 2 − ( d k ) 2 ) 2 = ( d 2 k 4 − d 2 k 2 ) 2 = ( d 2 k 4 − d 2 k 2 ) ( d 2 k 4 − d 2 k 2 ) = d 2 k 2 ( k 2 − 1 ) d 2 k 2 ( k 2 − 1 ) = d 4 k 4 ( k 2 − 1 ) 2 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a^2 - b^2)(c^2 - d^2) \\[1em] = [(dk^3)^2 - (dk^2)^2][(dk)^2 - (d^2)] \\[1em] = (d^2k^6 - d^2k^4)(d^2k^2 - d^2) \\[1em] = d^2k^4(k^2 - 1)d^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em] \text{R.H.S.} = (b^2 - c^2)^2 \\[1em] = ((dk^2)^2 - (dk)^2)^2 \\[1em] = (d^2k^4 - d^2k^2)^2 \\[1em] = (d^2k^4- d^2k^2)(d^2k^4 - d^2k^2) \\[1em] = d^2k^2(k^2 - 1)d^2k^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em] ∴ b a = c b = d c = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = bk = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( a 2 − b 2 ) ( c 2 − d 2 ) = [( d k 3 ) 2 − ( d k 2 ) 2 ] [( d k ) 2 − ( d 2 )] = ( d 2 k 6 − d 2 k 4 ) ( d 2 k 2 − d 2 ) = d 2 k 4 ( k 2 − 1 ) d 2 ( k 2 − 1 ) = d 4 k 4 ( k 2 − 1 ) 2 . R.H.S. = ( b 2 − c 2 ) 2 = (( d k 2 ) 2 − ( d k ) 2 ) 2 = ( d 2 k 4 − d 2 k 2 ) 2 = ( d 2 k 4 − d 2 k 2 ) ( d 2 k 4 − d 2 k 2 ) = d 2 k 2 ( k 2 − 1 ) d 2 k 2 ( k 2 − 1 ) = d 4 k 4 ( k 2 − 1 ) 2 .
Since, L.H.S. = R.H.S. hence, proved that,
(a2 - b2 )(c2 - d2 ) = (b2 - c2 )2 .
(iii) Since, a, b, c, d are in continued proportion
∴ a b = b c = c d = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = b k = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( a + d ) ( b + c ) − ( a + c ) ( b + d ) = ( d k 3 + d ) ( d k 2 + d k ) − ( d k 3 + d k ) ( d k 2 + d ) = d ( k 3 + 1 ) d k ( k + 1 ) − d k ( k 2 + 1 ) d ( k 2 + 1 ) = d 2 k [ ( k 3 + 1 ) ( k + 1 ) − ( k 2 + 1 ) 2 ] = d 2 k [ ( k 4 + k 3 + k + 1 ) − ( k 4 + 1 + 2 k 2 ) ] = d 2 k ( k 4 − k 4 + k 3 − 2 k 2 + k + 1 − 1 ) = d 2 k ( k 3 − 2 k 2 + k ) = d 2 k ( k ( k 2 − 2 k + 1 ) ) = d 2 k 2 ( k 2 − 2 k + 1 ) = d 2 k 2 ( k − 1 ) 2 . R.H.S. = ( b − c ) 2 = ( d k 2 − d k ) 2 = [ ( d k ) 2 ( k − 1 ) 2 ] = d 2 k 2 ( k − 1 ) 2 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a + d)(b + c) - (a + c)(b + d) \\[1em] = (dk^3 + d)(dk^2 + dk) - (dk^3 + dk)(dk^2 + d) \\[1em] = d(k^3 + 1)dk(k + 1) - dk(k^2 + 1)d(k^2 + 1) \\[1em] = d^2k[(k^3 + 1)(k + 1) - (k^2 + 1)^2] \\[1em] = d^2k[(k^4 + k^3 + k + 1) - (k^4 + 1 + 2k^2)] \\[1em] = d^2k(k^4 - k^4 + k^3 - 2k^2 + k + 1 - 1) \\[1em] = d^2k(k^3 - 2k^2 + k) \\[1em] = d^2k(k(k^2 - 2k + 1)) \\[1em] = d^2k^2(k^2 - 2k + 1) \\[1em] = d^2k^2(k - 1)^2. \\[1em] \text{R.H.S.} = (b - c)^2 \\[1em] = (dk^2 - dk)^2 \\[1em] = [(dk)^2(k - 1)^2] \\[1em] = d^2k^2(k - 1)^2. ∴ b a = c b = d c = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = bk = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( a + d ) ( b + c ) − ( a + c ) ( b + d ) = ( d k 3 + d ) ( d k 2 + d k ) − ( d k 3 + d k ) ( d k 2 + d ) = d ( k 3 + 1 ) d k ( k + 1 ) − d k ( k 2 + 1 ) d ( k 2 + 1 ) = d 2 k [( k 3 + 1 ) ( k + 1 ) − ( k 2 + 1 ) 2 ] = d 2 k [( k 4 + k 3 + k + 1 ) − ( k 4 + 1 + 2 k 2 )] = d 2 k ( k 4 − k 4 + k 3 − 2 k 2 + k + 1 − 1 ) = d 2 k ( k 3 − 2 k 2 + k ) = d 2 k ( k ( k 2 − 2 k + 1 )) = d 2 k 2 ( k 2 − 2 k + 1 ) = d 2 k 2 ( k − 1 ) 2 . R.H.S. = ( b − c ) 2 = ( d k 2 − d k ) 2 = [( d k ) 2 ( k − 1 ) 2 ] = d 2 k 2 ( k − 1 ) 2 .
Since, L.H.S = R.H.S, hence proved that,
(a + d)(b + c) - (a + c)(b + d) = (b - c)2 .
(iv) Since, a, b, c, d are in continued proportion
∴ a b = b c = c d = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = b k = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = a : d = a d = d k 3 d = k 3 . \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = a : d \\[1em] = \dfrac{a}{d} \\[1em] = \dfrac{dk^3}{d} \\[1em] = k^3. \\[1em] ∴ b a = c b = d c = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = bk = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = a : d = d a = d d k 3 = k 3 .
R.H.S. = triplicate ratio of (a - b) : (b - c)
= ( a − b ) 3 : ( b − c ) 3 = ( a − b ) 3 ( b − c ) 3 = ( d k 3 − d k 2 ) 3 ( d k 2 − d k ) 3 = [ k ( d k 2 − d k ) ] 3 ( d k 2 − d k ) 3 = k 3 ( d k 2 − d k ) 3 ( d k 2 − d k ) 3 = k 3 . = (a - b)^3 : (b - c)^3 \\[1em] = \dfrac{(a - b)^3}{(b - c)^3} \\[1em] = \dfrac{(dk^3 - dk^2)^3}{(dk^2 - dk)^3} \\[1em] = \dfrac{[k(dk^2 - dk)]^3}{(dk^2 - dk)^3} \\[1em] = \dfrac{k^3(dk^2 - dk)^3}{(dk^2 - dk)^3} \\[1em] = k^3. = ( a − b ) 3 : ( b − c ) 3 = ( b − c ) 3 ( a − b ) 3 = ( d k 2 − d k ) 3 ( d k 3 − d k 2 ) 3 = ( d k 2 − d k ) 3 [ k ( d k 2 − d k ) ] 3 = ( d k 2 − d k ) 3 k 3 ( d k 2 − d k ) 3 = k 3 .
Since, L.H.S. = R.H.S. hence proved that,
a : d = triplicate ratio of (a - b) : (b - c).
(v) Since, a, b, c, d are in continued proportion
∴ a b = b c = c d = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = b k = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( a − b c + a − c b ) 2 − ( d − b c + d − c b ) 2 . = ( d k 3 − d k 2 d k + d k 3 − d k 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 R.H.S. = ( a − d ) 2 ( 1 c 2 − 1 b 2 ) = ( d k 3 − d ) 2 ( 1 ( d k ) 2 − 1 ( d k 2 ) 2 ) = d 2 ( k 3 − 1 ) 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 \therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2. \\[1em] = \big(\dfrac{dk^3 - dk^2}{dk} + \dfrac{dk^3 - dk}{dk^2}\big)^2 - \big(\dfrac{d - dk^2}{dk} + \dfrac{d - dk}{dk^2}\big)^2. \\[1em] = \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] = \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] = \big(\dfrac{dk^4 - dk}{dk^2} \big)^2 - \big(\dfrac{d - dk^3}{dk^2}\big)^2. \\[1em] = \big(\dfrac{dk(k^3 - 1)}{dk^2}\big)^2 - \big(\dfrac{d(1 - k^3)}{dk^2}\big)^2 \\[1em] = \dfrac{d^2k^2(k^3 - 1)^2}{d^2k^4} - \dfrac{d^2(1 - k^3)^2}{d^2k^4} \\[1em] = \dfrac{(k^3 - 1)^2}{k^2} - \dfrac{(1 - k^3)^2}{k^4} \\[1em] = \dfrac{k^6 + 1 -2k^3}{k^2} - \dfrac{1 + k^6 - 2k^3}{k^4} \\[1em] = \dfrac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4} \\[1em] = \dfrac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4} \\[1em] \text{R.H.S.} = (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] = d^2(k^3 - 1)^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} ∴ b a = c b = d c = k ∴ c = d k , b = c k = ( d k ) k = d k 2 and a = bk = ( c k ) k = ( d k ) k 2 = d k 3 . L.H.S. = ( c a − b + b a − c ) 2 − ( c d − b + b d − c ) 2 . = ( d k d k 3 − d k 2 + d k 2 d k 3 − d k ) 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 R.H.S. = ( a − d ) 2 ( c 2 1 − b 2 1 ) = ( d k 3 − d ) 2 ( ( d k ) 2 1 − ( d k 2 ) 2 1 ) = d 2 ( k 3 − 1 ) 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 b ) 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}{b}\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 + b a − c ) 2 − ( c d − b + b d − c ) 2 = ( a − d ) 2 ( c 2 1 − b 2 1 ) .
