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If three unequal positive real numbers satisfying b^2=ac and b – c, c – a, a – b are in H.P. then the value of a + b + c is independent of |
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Answer» Given : If THREE unequal positive real numbers satisfying b² = ac and b – c, c – a, a – b are in H.P . To find : The VALUE of (a + b + c) is independent of solution : we have , b – c, c – a, a – b are in H.P . so, 1/(b - c) , 1/(c - a) and 1/(a - b) are in A.P ⇒1/(c - a) - 1/(b - c) = 1/(a - b) - 1/(c - a) ⇒2/(c - a) = 1/(b - c) + 1/(a - b) ⇒2/(c - a) = [(a - b) + (b - c)]/(b - c)(a - b) ⇒2/(c - a) = (a - c)/(b - c)(a - b) ⇒2(b - c)(a - b) = -(a - c)² ⇒2[ab - b² - ac + bc ] = - [a² + c² - 2ac ] ⇒2ab - 2b² - 2ac + 2bc = - a² - c² + 2ac ⇒a² + c² + 2ab - 2b² - 4AC + 2bc = 0 ⇒a² + b² + c² + 2ab + 2bc + 2ca - 6ca - 3b² = 0 ⇒(a + b + c)² - 6ca - 3b² = 0 ⇒(a + b + c)² - 6b² - 3b² = 0 [ as given b² = ca] ⇒(a + b + c)² = 9b² ⇒(a + b + c) = 3b Hence the value of (a + b + c) is independent of a and c. also read similar QUESTIONS : Three unequal positive numbers a, b, c are in gp then prove that a + c >2b If a, b, care in H.P., then value of c/a+ b-c/b-a +ab+bc/ac (C) 3 (D)4 (A) 1 (B) 2 |
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