For any three positive real numbers $\mathrm{a}, \mathrm{b}$ and c , $9\left(25 a^{2}+b^{2}\right)+25\left(c^{2}-3 a c\right)=15 b(3 a+c)$. Then :

(C) $b, c$ and $a$ are in A.P.

(A) $a, b$ and $c$ are in G.P.

(B) $b, c$ and $a$ are in G.P.

(D) $a, b$ and $c$ are in A.P.

JEE Main · 2017 · 02 Apr 2017 · Offline · Sequences and Series

For any three positive real numbers

a, b

and c ,

9 (25 a^{2} + b^{2}) + 25 (c^{2} - 3 a c) = 15 b (3 a + c)

.
Then :

Answer: (C)

Multiplying out gives

225 a^{2} + 9 b^{2} + 25 c^{2} - 45 ab - 15 b c - 75 a c = 0

. This is exactly

\frac{1}{2} [(15 a - 3 b)^{2} + (3 b - 5 c)^{2} + (15 a - 5 c)^{2}] = 0

, a sum of squares, so each bracket vanishes:

15 a = 3 b

3 b = 5 c

and

15 a = 5 c

. Hence

b = 5 a

and

c = 3 a

. Looking at

b, c, a

5 a, 3 a, a

, the successive differences are both

- 2 a

, so

b, c, a

are in arithmetic progression.

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option