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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 real numbers, then a,b,c are in GP.
Statement 2 If `(a_(1)-a_(2))^(2)+(a_(2)-a_(3))^(2)+(a_(3)-a_(1))^(2)=0`, then `a_(1)=a_(2)=a_(3),AA a_(1),a_(2),a_(3) in R`.

A

Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1

B

Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1

C

Statement 1 is true, Statement 2 is false

D

Statement 1 is false, Statement 2 is true

Text Solution

Verified by Experts

The correct Answer is:
D
Statement 1 `4a^(2)+9b^(2)+16c^(2)-2(3ab+6bc+4ca)=0`
`implies (2a)^(2)+(3b)^(2)+(4c)^(2)-(2a)(3b)-(3b)(4c)-(2a)(4c)=0`
`implies (1)/(2){(2a-3b)^(2)+(3b-4c)^(2)+(4c-2a)^(2)}=0`
`implies 2a-3b=0" and "3b-4c=0" and "4c-2a=0`
`implies" and " b=(4c)/(3)" and " c=(a)/(2) implies a=(3b)/(2)" and " b=(4c)/(3)" and " c=(3b)/(4)`
Then,a,b,c are of the form `(3b)/(2),b,(3b)/(4)`, which are in HP.
So, Statement 1 is false.
Statement 2 If `(a_(1)-a_(2))^(2)+(a_(2)-a_(3))^(2)+(a_(3)-a_(1))^(2)=0`
`implies a_(1)-a_(2)=0" and "a_(2)-a_(3)=0" and a_(3)-a_(1)=0`
`implies a_(1)=a_(2)=a_(3),AA a_(1),a_(2),a_(3) inR`
So, Statement 2 istrue.
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