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# If a,b,c are real and $\;4a^2+9b^2+16c^2-6ab-12bc-8ac=0\;$ the a,b,c are in

$(a)\;AP\qquad(b)\;GP\qquad(c)\;HP\qquad(d)\;None\;of\;these$

Explanation : Multiplying by 2 we get ,
$8a^2+18b^2+32c^2-12ab-24bc-16ac=0$
$(2a-3b)^2+(3b-4c)^2+(4c-2a)^2=0$
$2a=3b=4c$
a , b , c are in HP.