Question

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

• A. $18(a+b+c{)}^2=18\left(a^2+b^2+c^2\right)+ab$
• B. a, b, c are in GP.
• C. a, $2b,3c$ 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.
$\Rightarrow 2\log b=\log a+\log c$
$\therefore b^2=ac\dots .(1)$
$\Rightarrow a,b,c\text{ are in GP. }$
(B)also given $(\log a-\log 2b),(\log 2b-\log 3c),(\log 3c-\log a)$ are in A.P.
$\Rightarrow 2(\log 2b-\log 3c)=\log 3c-\log 2b$
$\Rightarrow 3\log 2b=3\log 3c$
$\therefore 2b=3c\dots (2)\Rightarrow 4b^2=9c^2$
from (1) and (3)
$4ac=9c^2\Rightarrow a=\frac{9c}{4}\text{ and }b=\frac{3c}{2}$
$a=\frac{9c}{4};b=\frac{3c}{2}\text{ and }c=c$
$\therefore a,b,c\text{ forms the sides of triangle }\Rightarrow (D)$
$a+b>c;b+c>a;c+a>b$
but $a,2b$ and $3c$ are not in H.P.