If a,b,c are in A.P., then `2^(ax+1),2^(bx+1),2^(cx+1), x in R`, are in

Options

A. A.P
B. G.P. and when `x gt 0`
C. G.P. only when `x lt 0`
D. G.P. for all x

Answer

To solve the problem, we need to show that if \( a, b, c \) are in arithmetic progression (A.P.), then the expressions \( 2^{ax+1}, 2^{bx+1}, 2^{cx+1} \) are in geometric progression (G.P.) for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( a, b, c \) are in A.P., we have the condition:
\[ 2b = a + c \]

2. **Expressing the Terms**:
We need to analyze the terms \( 2^{ax+1}, 2^{bx+1}, 2^{cx+1} \). We can rewrite these terms as:
\[ 2^{ax+1} = 2^{ax} \cdot 2^1 = 2^{ax} \cdot 2 \]
\[ 2^{bx+1} = 2^{bx} \cdot 2^1 = 2^{bx} \cdot 2 \]
\[ 2^{cx+1} = 2^{cx} \cdot 2^1 = 2^{cx} \cdot 2 \]
Thus, we can factor out \( 2 \):
\[ 2^{ax+1}, 2^{bx+1}, 2^{cx+1} \text{ can be expressed as } 2 \cdot 2^{ax}, 2 \cdot 2^{bx}, 2 \cdot 2^{cx} \]

3. **Finding the Ratios**:
To check if these terms are in G.P., we need to check the ratio of the second term to the first and the third term to the second:
\[ \text{Ratio } \frac{2^{bx+1}}{2^{ax+1}} = \frac{2 \cdot 2^{bx}}{2 \cdot 2^{ax}} = \frac{2^{bx}}{2^{ax}} = 2^{(b-a)x} \]
\[ \text{Ratio } \frac{2^{cx+1}}{2^{bx+1}} = \frac{2 \cdot 2^{cx}}{2 \cdot 2^{bx}} = \frac{2^{cx}}{2^{bx}} = 2^{(c-b)x} \]

4. **Using the A.P. Condition**:
Since \( a, b, c \) are in A.P., we have:
\[ b - a = c - b \]
Let \( d = b - a = c - b \). Then we can express:
\[ c - b = d \quad \text{and} \quad b - a = d \]

5. **Equating the Ratios**:
From the ratios we found:
\[ \frac{2^{bx+1}}{2^{ax+1}} = 2^{(b-a)x} = 2^{dx} \]
\[ \frac{2^{cx+1}}{2^{bx+1}} = 2^{(c-b)x} = 2^{dx} \]
Since both ratios are equal, we have:
\[ \frac{2^{bx+1}}{2^{ax+1}} = \frac{2^{cx+1}}{2^{bx+1}} \]

6. **Conclusion**:
Since the ratios are equal, we conclude that \( 2^{ax+1}, 2^{bx+1}, 2^{cx+1} \) are in G.P.

### Final Answer:
Thus, if \( a, b, c \) are in A.P., then \( 2^{ax+1}, 2^{bx+1}, 2^{cx+1} \) are in G.P.