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If 4a^(2)+9b^(2)+16c^(2)=2(3ab+6bc+4ca),...

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

A

AP

B

GP

C

HP

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
c

Since , ` 4a^(2) + 9b^(2) + 16c^(2) = 2(3ab + 6bc+4ca)`
`rArr 8a^(2) + 18b^(2) + 32c^(2) - 12ab - 24bc - 16ca = 0 `
`rArr (4a^(2)- 12ab + 9b^(2)-24bc +16c^(2))`
` + (16c^(2) - 16ca+4a^(2))^(2)=0`
` rArr ` 2a = 3b , 3b = 4c, 4c = 2a
` rArr` 2a = 3b = 4c = k [say]
` rArr a= k/2 , b = k/3, c = k/4 `
` rArr ` are in HP.
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