Let a, b, c > 1, a ^ 3 , b ^ 3 and c ^ 3 be in A . P . and _ a b ,...

Given,

${a}^{3}, {b}^{3}, {c}^{3}$ be in $A . P .$

So, ${a}^{3} + {c}^{3} = 2{b}^{3} . . . . . . . . \left(1\right)$

Also given $\log_{a}b, \log_{c}a \text{ and } \log_{b}c$ are in $G . P .$

So, $\dfrac{logb}{loga} \cdot \dfrac{logc}{logb} = {\left(\dfrac{loga}{logc}\right)}^{2}$

$⇒ ( loga{)}^{3} = ( logc{)}^{3} ⇒ a = c$ ..........(2)

Now from equation \left(1\right) \text{ and } \left(2\right)$we get,$a = b = c$</p><p>Now,${T}_{1} = \dfrac{a + 4b + c}{3} = 2a and $d = \dfrac{a - 8b + c}{10} = \dfrac{- 6a}{10} = \dfrac{- 3}{5}a$

$∴ {S}_{20} = \dfrac{20}{2}\left[4a + 19\left(- \dfrac{3}{5}a\right)\right]$

$⇒ {S}_{20} = 10\left[\dfrac{20a - 57a}{5}\right]$

$⇒ {S}_{20} = - 74a$

$⇒ - 444 = - 74a ⇒ a = 6$

Hence, $abc = {6}^{3} = 216$