Let a , b , c > 1 , a 3 , b 3 and c 3 be in A . P . and log a b , log c a and log b c be in G . P . If…

Given,

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

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

Also given ${\log }_{a}b, {\log }_{c}a \& {\log }_{b}c$  are in $\mathrm{G}.\mathrm{P}.$

So, $\frac{\log b}{\log a}·\frac{\log c}{\log b}={\left(\frac{\log a}{\log c}\right)}^2$

$\Rightarrow (\log a{)}^3=(\log c{)}^3\Rightarrow a=c$  ..........(2)

Now from equation $\left(1\right) \& \left(2\right)$ we get, $a=b=c$

Now, $T_1=\frac{a+4b+c}{3}=2a$ and $d=\frac{a-8b+c}{10}=\frac{-6a}{10}=\frac{-3}{5}\mathrm{a}$

$\therefore S_{20}=\frac{20}{2}\left[4a+19\left(-\frac{3}{5}\mathrm{a}\right)\right]$

$\Rightarrow S_{20}=10\left[\frac{20a-57a}{5}\right]$

$\Rightarrow S_{20}=-74a$

$\Rightarrow -444=-74\mathrm{a}\Rightarrow \mathrm{a}=6$

Hence, $abc=6^3=216$