Question

If the triplets $\log a, \log b, \log c$ and $(\log a-\log 2 b),(\log 2 b-\log 3 c),(\log 3 c-\log a)$ are in arithmetic progression then

• A. $18(a+b+c)^2=18\left(a^2+b^2+c^2\right)+a b$
• B. a, b, c are in GP.
• C. a, $2 b , 3 c$ are in H.P.
• D. a, b, c can be the lengths of the sides of a triangle

Answer 1

Answer: (B), (D)

$\log a, \log b, \log c$ are in A.P.
$\Rightarrow 2 \log b =\log a +\log c $
$\therefore b ^2= ac \ldots .(1) $
$\Rightarrow a , b , c \text { are in GP. }$
(B)also given $(\log a-\log 2 b),(\log 2 b-\log 3 c),(\log 3 c-\log a)$ are in A.P.
$\Rightarrow 2(\log 2 b-\log 3 c)=\log 3 c-\log 2 b $
$\Rightarrow 3 \log 2 b=3 \log 3 c$
$\therefore 2 b=3 c \ldots(2) \Rightarrow 4 b^2=9 c^2$
from (1) and (3)
$4 a c=9 c^2 \Rightarrow a=\frac{9 c}{4} \text { and } b=\frac{3 c}{2}$
$ a =\frac{9 c }{4} ; b =\frac{3 c }{2} \text { and } c = c $
$\therefore a , b , c \text { forms the sides of triangle } \Rightarrow(D)$
$ a + b > c ; b + c > a ; c + a > b$
but $a, 2 b$ and $3 c$ are not in H.P.