If _ e a, _ e b, _ e c are in an A . P . and _ e a - _ e 2b, _ e 2b...

Given: $loga, logb, logc$ are in $A . P .$

$⇒ 2logb = loga + logc$

$⇒ {b}^{2} = ac . . . \left(1\right)$

And $loga - \log 2b, \log 2b - \log 3c, \log 3c - loga$ are in $A . P$ 

$⇒ 2log\dfrac{2b}{3c} = log\dfrac{a}{2b} + log\dfrac{3c}{a}$

$⇒ {\left(\dfrac{2b}{3c}\right)}^{2} = \dfrac{a}{2b} \times \dfrac{3c}{a}$

$⇒ {\left(\dfrac{2b}{3c}\right)}^{2} = \dfrac{3c}{2b}$

$⇒ 2b = 3c$ 

$⇒ 4{b}^{2} = 9{c}^{2} \text{ and } 6b = 9c . . . \left(2\right)$

Now solving the equation \left(1\right) \text{ and } \left(2\right)$we get,</p><p>$4ac = 9{c}^{2}$</p><p>$⇒ 4a = 9c . . . \left(3\right)$</p><p>Then from equation \left(2\right) \text{ and } \left(3\right)$ we get,

$4a = 6b = 9c = k$

$⇒ a = \dfrac{k}{4}, b = \dfrac{k}{6} \text{ and } c = \dfrac{k}{9}$

$⇒ a: b: c = \dfrac{1}{4}: \dfrac{1}{6}: \dfrac{1}{9} = 9: 6: 4$