JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q68

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MathematicsSingle correct+4 / -1Question 68 of 90

For any three positive real numbers a, b and c,
$9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c).$
Then :

b, c and a are in A.P.

a, b and c are in A.P.

a, b and c are in G.P.

b, c and a are in G.P.

Answer and worked solution

Official answer: A

With $x=15a$ , $y=3b$ , $z=5c$ the condition is $x^2+y^2+z^2-xy-yz-zx=0$ , which forces $x=y=z$ . So $b=5a$ and $c=3a$ , and b, c, a $=5a,3a,a$ are in A.P.

Topics:Arithmetic progression Inequalities and modulus

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 68