1. Given conditions

We have:

• $a,b,c>1$
• $a^3,b^3,c^3$ are in A.P.
• $\log_a b,\log_c a,\log_b c$ are in G.P.

Also, an A.P. has:

• first term $A=\dfrac{a+4b+c}{3}$
• common difference $d=\dfrac{a-8b+c}{10}$
• sum of first $20$ terms is $-444$

We need to find $abc$.

2. Use the A.P. condition on cubes

Since $a^3,b^3,c^3$ are in A.P., 2b^3=a^3+c^3. \tag{1}

3. Use the G.P. condition on logarithms

Let $x=\log_a b,\quad y=\log_c a,\quad z=\log_b c.$

Given $x,y,z$ are in G.P., so y^2=xz. \tag{2}

Now write these in natural logs: $x=\frac{\ln b}{\ln a},\quad y=\frac{\ln a}{\ln c},\quad z=\frac{\ln c}{\ln b}.$

Then $xz=\frac{\ln b}{\ln a}\cdot\frac{\ln c}{\ln b}=\frac{\ln c}{\ln a}.$

And $y^2=\left(\frac{\ln a}{\ln c}\right)^2.$

So from $y^2=xz$, $\left(\frac{\ln a}{\ln c}\right)^2=\frac{\ln c}{\ln a}.$

Let $r=\frac{\ln a}{\ln c}>0.$ Then $r^2=\frac{1}{r} \implies r^3=1.$ Since $r>0$, we get r=1 \implies \ln a=\ln c \implies a=c. \tag{3}

4. Combine with the cube A.P. condition

Using $a=c$ in (1): $2b^3=a^3+a^3=2a^3$ $\implies b^3=a^3$ Since $a,b>1$, this gives $a=b.$ And with $a=c$, we obtain a=b=c. \tag{4}

5. Use the sum of 20 terms

Since $a=b=c$, let $a=b=c=t.$

Then the first term becomes $A=\frac{t+4t+t}{3}=\frac{6t}{3}=2t.$

The common difference becomes $d=\frac{t-8t+t}{10}=\frac{-6t}{10}=-\frac{3t}{5}.$

Now sum of first $20$ terms: $S_{20}=\frac{20}{2}\left[2A+(20-1)d\right].$ So $-444=10\left[2(2t)+19\left(-\frac{3t}{5}\right)\right].$

Simplify inside: $2(2t)=4t,$ $19\left(-\frac{3t}{5}\right)=-\frac{57t}{5}.$ Hence $-444=10\left(4t-\frac{57t}{5}\right).$

Now $4t=\frac{20t}{5},$ so $4t-\frac{57t}{5}=\frac{20t-57t}{5}=-\frac{37t}{5}.$

Thus $-444=10\left(-\frac{37t}{5}\right)=-74t.$ Therefore $t=6.$

So $a=b=c=6.$

6. Find $abc$

$abc=6\cdot 6\cdot 6=216.$

7. Check options

• A: $343$ ❌
• B: $216$ ✅
• C: $\dfrac{343}{8}$ ❌
• D: $\dfrac{125}{8}$ ❌

Therefore, the correct option is B.