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 the sum of first `20` terms of an A.P.,whose first terms is `(a+4b+c)/(3)` and the common difference is `(a-8b+c)/(10)` is `-444`,then abc is equal to:

Options

A. `343`
B. `216`
C. `(343)/(8)`
D. `(125)/(8)`

Answer

To solve the problem step by step, we will break down the given conditions and derive the required value of \( abc \).

### Step 1: Understanding the A.P. condition
We know that \( a^3, b^3, c^3 \) are in A.P. This means:
\[ 2b^3 = a^3 + c^3 \]

### Step 2: Using the property of logarithms
We also know that \( \log_a b, \log_c a, \log_b c \) are in G.P. This means:
\[ (\log_c a)^2 = \log_a b \cdot \log_b c \]
Using the change of base formula, we can express these logarithms in terms of natural logarithms:
\[ \log_a b = \frac{\log b}{\log a}, \quad \log_c a = \frac{\log a}{\log c}, \quad \log_b c = \frac{\log c}{\log b} \]

### Step 3: Setting up the equation from G.P.
Substituting these into the G.P. condition:
\[ \left(\frac{\log a}{\log c}\right)^2 = \left(\frac{\log b}{\log a}\right) \left(\frac{\log c}{\log b}\right) \]
This simplifies to:
\[ \frac{(\log a)^2}{(\log c)^2} = \frac{\log c}{\log a} \]
Cross-multiplying gives:
\[ (\log a)^3 = (\log c)^3 \]
Thus, we conclude:
\[ \log a = \log c \implies a = c \]

### Step 4: Substituting \( c \) with \( a \)
Since \( a = c \), we can substitute \( c \) in the A.P. condition:
\[ 2b^3 = a^3 + a^3 \implies 2b^3 = 2a^3 \implies b^3 = a^3 \implies b = a \]

### Step 5: Conclusion about \( a, b, c \)
From the above steps, we have:
\[ a = b = c \]

### Step 6: Finding the first term and common difference
The first term of the A.P. is given by:
\[ \frac{a + 4b + c}{3} = \frac{a + 4a + a}{3} = \frac{6a}{3} = 2a \]
The common difference is given by:
\[ \frac{a - 8b + c}{10} = \frac{a - 8a + a}{10} = \frac{-6a}{10} = -\frac{3a}{5} \]

### Step 7: Using the sum of the first 20 terms
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \left(2(2a) + (20-1)(-\frac{3a}{5})\right) = 10 \left(4a - \frac{57a}{5}\right) \]
Calculating this gives:
\[ S_{20} = 10 \left(\frac{20a - 57a}{5}\right) = 10 \left(-\frac{37a}{5}\right) = -74a \]
We are given that \( S_{20} = -444 \):
\[ -74a = -444 \implies a = \frac{444}{74} = 6 \]

### Step 8: Finding \( abc \)
Since \( a = b = c = 6 \):
\[ abc = 6 \cdot 6 \cdot 6 = 216 \]

### Final Answer
Thus, the value of \( abc \) is:
\[ \boxed{216} \]