Let a,b,c be in AP and `k ne 0` be a real number. Which of following correct ?
(1) ka, kb, kc are in AP
(2) `k-a,k-b,k-c`
(3) `(a)/(k),(b)/(k),(c)/(k)` are in AP
Select the correct answer using the given code below:

Options

A. A)1 and 2 only
B. B)2 and 3 only
C. C)1 and 3 only
D. D)1,2 and 3

Answer

To solve the problem, we need to determine whether the given statements about the sequences formed by \( a, b, c \) (which are in arithmetic progression) are correct.

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression (AP)**:
- If \( a, b, c \) are in AP, then \( 2b = a + c \). This means that the middle term \( b \) is the average of \( a \) and \( c \).

2. **Statement 1: \( ka, kb, kc \) are in AP**:
- To check if \( ka, kb, kc \) are in AP, we need to verify if \( 2(kb) = ka + kc \).
- Substituting \( kb \) gives us:
\[ 2(kb) = 2(k \cdot b) = k(2b) \]
\[ ka + kc = k(a + c) \]
- Since \( 2b = a + c \), we have:
\[ k(2b) = k(a + c) \]
- Therefore, \( ka, kb, kc \) are in AP.

3. **Statement 2: \( k-a, k-b, k-c \)**:
- To check if \( k-a, k-b, k-c \) are in AP, we need to verify if \( 2(k-b) = (k-a) + (k-c) \).
- This simplifies to:
\[ 2(k-b) = 2k - 2b \]
\[ (k-a) + (k-c) = k - a + k - c = 2k - (a + c) \]
- Since \( a + c = 2b \) (from the definition of AP), we have:
\[ 2k - (a + c) = 2k - 2b \]
- Thus, \( k-a, k-b, k-c \) are also in AP.

4. **Statement 3: \( \frac{a}{k}, \frac{b}{k}, \frac{c}{k} \)**:
- To check if \( \frac{a}{k}, \frac{b}{k}, \frac{c}{k} \) are in AP, we need to verify if \( 2\left(\frac{b}{k}\right) = \frac{a}{k} + \frac{c}{k} \).
- This simplifies to:
\[ 2\left(\frac{b}{k}\right) = \frac{2b}{k} \]
\[ \frac{a}{k} + \frac{c}{k} = \frac{a + c}{k} \]
- Since \( a + c = 2b \), we have:
\[ \frac{a + c}{k} = \frac{2b}{k} \]
- Thus, \( \frac{a}{k}, \frac{b}{k}, \frac{c}{k} \) are also in AP.

### Conclusion:
All three statements are correct. Therefore, the correct answer is option D: 1, 2, and 3.