Which of the following sequqnces are in arithmetic progression?
(i) 2,6,10,14,…. (ii) 15,12,9,6
(iii) 5,9,12,18,….. (iv) `1/2, 1/3, 1/4, 1/5,….`

Answer

To determine which of the given sequences are in arithmetic progression (AP), we need to check if the difference between consecutive terms is constant for each sequence.

### Step-by-Step Solution:

1. **Check the first sequence: 2, 6, 10, 14**
- Let the terms be:
- \( A_1 = 2 \)
- \( A_2 = 6 \)
- \( A_3 = 10 \)
- \( A_4 = 14 \)
- Calculate the common difference \( D \):
- \( D_1 = A_2 - A_1 = 6 - 2 = 4 \)
- \( D_2 = A_3 - A_2 = 10 - 6 = 4 \)
- \( D_3 = A_4 - A_3 = 14 - 10 = 4 \)
- Since \( D_1 = D_2 = D_3 = 4 \), this sequence is in AP.

2. **Check the second sequence: 15, 12, 9, 6**
- Let the terms be:
- \( A_1 = 15 \)
- \( A_2 = 12 \)
- \( A_3 = 9 \)
- \( A_4 = 6 \)
- Calculate the common difference \( D \):
- \( D_1 = A_2 - A_1 = 12 - 15 = -3 \)
- \( D_2 = A_3 - A_2 = 9 - 12 = -3 \)
- \( D_3 = A_4 - A_3 = 6 - 9 = -3 \)
- Since \( D_1 = D_2 = D_3 = -3 \), this sequence is also in AP.

3. **Check the third sequence: 5, 9, 12, 18**
- Let the terms be:
- \( A_1 = 5 \)
- \( A_2 = 9 \)
- \( A_3 = 12 \)
- \( A_4 = 18 \)
- Calculate the common difference \( D \):
- \( D_1 = A_2 - A_1 = 9 - 5 = 4 \)
- \( D_2 = A_3 - A_2 = 12 - 9 = 3 \)
- \( D_3 = A_4 - A_3 = 18 - 12 = 6 \)
- Since \( D_1 \neq D_2 \neq D_3 \), this sequence is not in AP.

4. **Check the fourth sequence: \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \)**
- Let the terms be:
- \( A_1 = \frac{1}{2} \)
- \( A_2 = \frac{1}{3} \)
- \( A_3 = \frac{1}{4} \)
- \( A_4 = \frac{1}{5} \)
- Calculate the common difference \( D \):
- \( D_1 = A_2 - A_1 = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \)
- \( D_2 = A_3 - A_2 = \frac{1}{4} - \frac{1}{3} = \frac{3 - 4}{12} = -\frac{1}{12} \)
- Since \( D_1 \neq D_2 \), this sequence is not in AP.

### Summary of Results:
- Sequence (i): **In AP**
- Sequence (ii): **In AP**
- Sequence (iii): **Not in AP**
- Sequence (iv): **Not in AP**