Let `a_1=0` and `a_1,a_2,a_3` …. , `a_n` be real numbers such that `|a_i|=|a_(i-1) + 1|` for all I then the A.M. Of the number `a_1,a_2 ,a_3`…., `a_n` has the value A where : (a) `A lt -1/2` (b) `A lt -1` (c) `A ge -1/2` (d) A=-2

Options

A. `A lt -1/2`
B. `A lt -1`
C. `A ge -1/2`
D. A=-2

Answer

To solve the problem step by step, we start with the given conditions and derive the necessary values:

### Step 1: Understand the given information
We have:
- \( a_1 = 0 \)
- The condition \( |a_i| = |a_{i-1} + 1| \) for all \( i \).

### Step 2: Analyze the absolute value condition
The absolute value condition \( |a_i| = |a_{i-1} + 1| \) means that \( a_i \) can take on two possible forms based on the value of \( a_{i-1} \):
1. \( a_i = a_{i-1} + 1 \)
2. \( a_i = - (a_{i-1} + 1) \)

### Step 3: Explore the first case (positive values)
Assume \( a_i = a_{i-1} + 1 \):
- \( a_1 = 0 \)
- \( a_2 = 0 + 1 = 1 \)
- \( a_3 = 1 + 1 = 2 \)
- Continuing this way, we find:
\[ a_n = n - 1 \]

### Step 4: Calculate the sum of the sequence
The sum of the sequence \( a_1, a_2, \ldots, a_n \):
\[ S_n = 0 + 1 + 2 + \ldots + (n - 1) = \frac{(n - 1)n}{2} \]

### Step 5: Calculate the Arithmetic Mean (A.M.)
The arithmetic mean \( A \) is given by:
\[ A = \frac{S_n}{n} = \frac{\frac{(n - 1)n}{2}}{n} = \frac{n - 1}{2} \]

### Step 6: Explore the second case (negative values)
Now assume \( a_i = - (a_{i-1} + 1) \):
- \( a_1 = 0 \)
- \( a_2 = - (0 + 1) = -1 \)
- \( a_3 = - (-1 + 1) = 0 \)
- \( a_4 = - (0 + 1) = -1 \)
- Continuing this way, we see that:
- If \( n \) is even, the sequence will have equal numbers of \( -1 \) and \( 0 \).
- If \( n \) is odd, the last term will be \( 0 \).

### Step 7: Calculate the sum for the second case
For even \( n \):
\[ S_n = -1 \cdot \frac{n}{2} + 0 \cdot \frac{n}{2} = -\frac{n}{2} \]
For odd \( n \):
\[ S_n = -1 \cdot \frac{n-1}{2} + 0 \cdot \frac{n+1}{2} = -\frac{n-1}{2} \]

### Step 8: Calculate the Arithmetic Mean for the second case
For even \( n \):
\[ A = \frac{-\frac{n}{2}}{n} = -\frac{1}{2} \]
For odd \( n \):
\[ A = \frac{-\frac{n-1}{2}}{n} = -\frac{n-1}{2n} \]

### Step 9: Determine the minimum value of A
From the second case, we see that:
- For even \( n \), \( A = -\frac{1}{2} \).
- For odd \( n \), \( A \) approaches \( -\frac{1}{2} \) as \( n \) increases.

### Conclusion
Thus, in both cases, we find that:
\[ A \geq -\frac{1}{2} \]
The correct answer is:
**(c) \( A \geq -\frac{1}{2} \)**.