Sequences And Series Ques 37
For any three positive real numbers $a, b$ and $c$, if $9\left(25 a^{2}+b^{2}\right)+25\left(c^{2}-3 a c\right)=15 b(3 a+c)$, then(2017 Main)
(a) $b, c$ and $a$ are in GP
(b) $b, c$ and $a$ are in AP
(c) $a, b$ and $c$ are in $AP$
(d) $a, b$ and $c$ are in GP
Show Answer
Answer:
Correct Answer: 37.(b)
Solution:
- We have,
$225 a^{2}+9 b^{2}+25 c^{2}-75 a c-45 a b-15 b c=0 $
$\Rightarrow(15 a)^{2}+(3 b)^{2}+(5 c)^{2}-(15 a)(5 c)-(15 a)(3 b)-(3 b)(5 c)=0 $
$\Rightarrow \frac{1}{2}\left[(15 a-3 b)^{2}+(3 b-5 c)^{2}+(5 c-15 a)^{2}\right]=0$
$\Rightarrow 15 a=3 b, 3 b=5 c$ and $5 c=15 a$
$[\therefore 15 a=3 b=5 c]$
$\Rightarrow \frac{a}{1}=\frac{b}{5}=\frac{c}{3}=\lambda$
$\Rightarrow \quad a=\lambda, b=5 \lambda, c=3 \lambda$
$\therefore b, c, a$ are in AP.
BETA
