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If 4a 2 + 9b 2 + 16c 2 =2(3ab + 6bc + 4ca), where a,b,c are non-zero numbers, then a,b,c are in AP GP HP none of these

If 4a2 + 9b2 + 16c2 =2(3ab + 6bc + 4ca), where a,b,c are non-zero numbers, then a,b,c are in



  1. AP

  2. GP

  3. HP

  4. none of these

Grade:12

2 Answers

Amit Jain
askIITians Faculty 32 Points
8 years ago

167-1156_Sequence--Series.PNG

Amit Jain
IITD
AskIITIans
disha sharma
29 Points
9 months ago
Multiply by 2 on both sides
4a2+4a2+9b2+9b2+16c2+16c2–12ab–24bc–16ca = 0 ⇒ 
(2a–3b)2+(3b–4c)2+(4c–2a)2=0  ⇒ 
2a = 3b = 4c =
 λ ⇒ a = λ/2, b = λ/3 , c = λ/4 2,3,4 are in AP
⇒ 1/2, 1/3, 1/4 are in H.P. ⇒ λ/2, λ/3,λ/4 are in HP gives a, b, c are in HP -4a-2-9b-2-16c-2-2-3ab-6bc-4ca-where-a-b-c-are-non-zero-real-numbers-then-a-b-c-are-in hp

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