Question

Let $a, b, c>, a^3, b^3$ and $c^3$ be in AP, and $\log _a b, \log _c a$ and $\log _b c$ be in GP If the sum of first 20 terms of an AP, whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ is equal to:

• A. 343
• B. 216
• C. $\frac{343}{8}$
• D. $\frac{125}{8}$

Answer 1

The Correct Option is B

To solve the given problem, we need to find the value of \(abc\) based on the provided conditions. We have two main conditions here:

1. The numbers \(a^3\), \(b^3\), and \(c^3\) are in Arithmetic Progression (AP).
2. The logarithms \(\log_a b\), \(\log_c a\), and \(\log_b c\) are in Geometric Progression (GP).

Let's break down each part of the question:

Step 1: Understanding the AP Condition

Since \(a^3\), \(b^3\), and \(c^3\) are in AP, the condition for AP implies:

\(2b^3 = a^3 + c^3.\)

Step 2: Understanding the GP Condition

The logarithms \(\log_a b\), \(\log_c a\), and \(\log_b c\) are in GP. For three numbers to be in GP, the square of the middle term should be equal to the product of the other two terms:

\((\log_c a)^2 = \log_a b \cdot \log_b c.\)

Using the properties of logarithms, we transform the terms:

\(\log_a b = \frac{\log b}{\log a}, \, \log_c a = \frac{\log a}{\log c}, \, \log_b c = \frac{\log c}{\log b}.\)

Thus the GP condition transforms to:

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

Simplifying gives us:

\((\log a)^2 = \log b \cdot \log c.\)

Step 3: Solve the AP for 20 Terms

The first term of the new AP is given by:

\(A_1 = \frac{a+4b+c}{3},\)

and the common difference is:

\(d = \frac{a-8b+c}{10}.\)

The sum of the first 20 terms \(S_{20}\) of an AP is given by the formula:

\(S_n = \frac{n}{2} \cdot (2A_1 + (n-1)d).\)

Substituting the given values, we have:

\(S_{20} = \frac{20}{2} \cdot \left(2 \cdot \frac{a+4b+c}{3} + 19 \cdot \frac{a-8b+c}{10}\right) = -444.\)

This simplifies to:

\(10 \cdot \left(\frac{2(a+4b+c)}{3} + \frac{19(a-8b+c)}{10}\right) = -444.\)

Solving for \((a, b, c)\), we further simplify and equate coefficients as guided by the constraints of AP and GP:

\(a^3 + c^3 = 2b^3 \quad \text{and, after simplification,} \quad a = b = c.\)

Conclusion: Calculate \(abc\)

If we set \(a = b = c\), then their cubes and logarithms satisfy the given conditions regarding AP and GP. Thus:

\(a = b = c = 6 \rightarrow abc = 6 \cdot 6 \cdot 6 = 216.\)

Thus, the value of \(abc\) is 216.