Question

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

• A. $b$, $c$ and $a$ are in $GP$
• B. $b$ ,$c$ and $a$ are in $AP$
• C. $a$, $b$ and $c$ are in $HP$
• D. $a$, $b$ and $c$ are in $AGP$

Answer 1

The correct option is B $b$ ,$c$ and $a$ are in $AP$

(15a – 3b)^2 + (15a – 5c)^2+ (3b – 5c)^2 = 0

Let, 15a = 3b = 5c = 45λ

a = 3λ; b = 15λ; c = 9λ

2c = a + b

b, c, a are in A.P.