In an A.P., `T_(1) = log a, T_(n+1) = log b, T_(2n + 1) = log c`, then a, b, c are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none

Answer

To solve the problem, we need to establish the relationship between the terms in the arithmetic progression (A.P.) and how they relate to the values of \( a \), \( b \), and \( c \).

### Step-by-Step Solution:

1. **Identify the Terms in A.P.:**
We are given:
- \( T_1 = \log a \)
- \( T_{n+1} = \log b \)
- \( T_{2n+1} = \log c \)

2. **Using the A.P. Formula:**
The general term of an A.P. can be expressed as:
\[ T_k = T_1 + (k-1)D \]
where \( D \) is the common difference.

3. **Express \( T_{n+1} \):**
For \( T_{n+1} \):
\[ T_{n+1} = T_1 + nD \]
Substituting the values:
\[ \log b = \log a + nD \]
Rearranging gives:
\[ nD = \log b - \log a \]
Thus, we can express \( D \) as:
\[ D = \frac{\log b - \log a}{n} \]

4. **Express \( T_{2n+1} \):**
For \( T_{2n+1} \):
\[ T_{2n+1} = T_1 + (2n)D \]
Substituting the values:
\[ \log c = \log a + 2nD \]
Now substituting the expression for \( D \):
\[ \log c = \log a + 2n \left(\frac{\log b - \log a}{n}\right) \]
Simplifying this gives:
\[ \log c = \log a + 2(\log b - \log a) \]
\[ \log c = \log a + 2\log b - 2\log a \]
\[ \log c = 2\log b - \log a \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ \log c + \log a = 2\log b \]
This can be rewritten using properties of logarithms:
\[ \log(ac) = \log(b^2) \]
Thus, we have:
\[ ac = b^2 \]

6. **Conclusion:**
The relationship \( ac = b^2 \) indicates that \( a \), \( b \), and \( c \) are in a geometric progression (G.P.).

### Final Answer:
Thus, \( a, b, c \) are in G.P.

---