Permutations And Combinations

Question
CBSEENMA11015541

For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then

  • b , c and a are in G.P

  • b, c and a are in A.P

  • a, b and c are in A.P

  • a, b and c are in G.P

Solution

B.

b, c and a are in A.P

225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
(15a)2 + (3b)2 + (5c)2 – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP

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