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Question

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

A
AP
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B
GP
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C
HP
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D
none of these
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Solution

## The correct option is C HP4a2+9b2+16c2=4(3ab+6bc+4ca)⇒(4a2−12ab+9b2)+(9b2−24bc+16c2)+(16c2−16ca+4a2)=0⇒(2a−3b)2+(3b−4c)2+(4c−2a)2=0Therefore, 2a−3b=0; 3b−4c=0; 4c−2a=0⇒2a=3b=4c=k⇒a=k2; b=k3; c=k4now 2b=6k and 1c+1a=6k1c+1a=2bTherefore, a,b,c are in H.P

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