Question

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 AM of the numbers $a_1, a_2, a_3,......, a_n$ has the value $A$ where:

• A. $A < -\frac{1}{2}$
• B. $A < -1$
• C. $A \ge -\frac{1}{2}$
• D. $A = -\frac{1}{2}$

Answer 1

Answer: (C)

A \ge -\frac{1}{2}

Answer 2

The given conditions are:

1. $a_1 = 0$
2. $|a_i| = |a_{i-1} + 1|$ for all $i \ge 2$.

From the second condition, we can square both sides: $a_i^2 = (a_{i-1} + 1)^2$ $a_i^2 = a_{i-1}^2 + 2a_{i-1} + 1$

This relation holds for $i=2, 3, \ldots, n$. Let's sum this equation from $i=2$ to $n$: $\sum_{i=2}^{n} a_i^2 = \sum_{i=2}^{n} (a_{i-1}^2 + 2a_{i-1} + 1)$ $\sum_{i=2}^{n} a_i^2 = \sum_{i=2}^{n} a_{i-1}^2 + 2 \sum_{i=2}^{n} a_{i-1} + \sum_{i=2}^{n} 1$

Let $j = i-1$. Then as $i$ goes from $2$ to $n$, $j$ goes from $1$ to $n-1$. $\sum_{i=2}^{n} a_i^2 = \sum_{j=1}^{n-1} a_j^2 + 2 \sum_{j=1}^{n-1} a_j + (n-1)$

We can write $\sum_{i=2}^{n} a_i^2 = a_n^2 + \sum_{i=2}^{n-1} a_i^2$. And $\sum_{j=1}^{n-1} a_j^2 = a_1^2 + \sum_{j=2}^{n-1} a_j^2$. Substituting these into the equation: $a_n^2 + \sum_{i=2}^{n-1} a_i^2 = a_1^2 + \sum_{j=2}^{n-1} a_j^2 + 2 \sum_{j=1}^{n-1} a_j + (n-1)$ The summation terms on both sides cancel out: $a_n^2 = a_1^2 + 2 \sum_{j=1}^{n-1} a_j + (n-1)$

Since $a_1 = 0$, the equation simplifies to: $a_n^2 = 0^2 + 2 \sum_{j=1}^{n-1} a_j + (n-1)$ $a_n^2 = 2 \sum_{j=1}^{n-1} a_j + n-1$

Let $S_{n-1} = \sum_{j=1}^{n-1} a_j$. So, $a_n^2 = 2 S_{n-1} + n-1$. We are interested in the arithmetic mean $A = \frac{1}{n} \sum_{i=1}^{n} a_i = \frac{S_n}{n}$. We know $S_n = S_{n-1} + a_n$. From $a_n^2 = 2 S_{n-1} + n-1$, we can express $S_{n-1}$: $S_{n-1} = \frac{a_n^2 - (n-1)}{2}$.

Now substitute $S_{n-1}$ into the expression for $S_n$: $S_n = \frac{a_n^2 - (n-1)}{2} + a_n$ $S_n = \frac{a_n^2 - n + 1 + 2a_n}{2}$ $S_n = \frac{a_n^2 + 2a_n + 1 - n}{2}$ $S_n = \frac{(a_n+1)^2 - n}{2}$

Finally, the arithmetic mean $A$ is: $A = \frac{S_n}{n} = \frac{(a_n+1)^2 - n}{2n}$

We know that $(a_n+1)^2 \ge 0$ for any real number $a_n$. Therefore, $A \ge \frac{0 - n}{2n}$ $A \ge -\frac{n}{2n}$ $A \ge -\frac{1}{2}$

This shows that the arithmetic mean $A$ is always greater than or equal to $-\frac{1}{2}$.

To confirm that $A = -\frac{1}{2}$ is achievable, consider the sequence where $a_i = 0$ if $i$ is odd, and $a_i = -1$ if $i$ is even. Let's check if this sequence satisfies the condition $|a_i| = |a_{i-1} + 1|$:

• If $i$ is odd, $a_i = 0$. $a_{i-1} = -1$ (since $i-1$ is even). $|a_i| = |0| = 0$. $|a_{i-1} + 1| = |-1 + 1| = |0| = 0$. This holds.
• If $i$ is even, $a_i = -1$. $a_{i-1} = 0$ (since $i-1$ is odd). $|a_i| = |-1| = 1$. $|a_{i-1} + 1| = |0 + 1| = |1| = 1$. This holds.

Also, $a_1 = 0$, which is the starting condition.

Now, let's calculate the AM for this sequence: If $n$ is even, say $n=2k$: The sequence is $0, -1, 0, -1, \ldots, 0, -1$. There are $k$ terms of $0$ and $k$ terms of $-1$. Sum $S_n = k \times 0 + k \times (-1) = -k = -n/2$. $A = \frac{S_n}{n} = \frac{-n/2}{n} = -\frac{1}{2}$.

If $n$ is odd, say $n=2k+1$: The sequence is $0, -1, 0, -1, \ldots, -1, 0$. There are $k+1$ terms of $0$ and $k$ terms of $-1$. Sum $S_n = (k+1) \times 0 + k \times (-1) = -k = -(n-1)/2$. $A = \frac{S_n}{n} = \frac{-(n-1)/2}{n} = -\frac{n-1}{2n} = -\frac{1}{2} + \frac{1}{2n}$.

Since $A$ can be exactly $-\frac{1}{2}$ (when $n$ is even) and $A$ is always greater than or equal to $-\frac{1}{2}$, the correct option is $A \ge -\frac{1}{2}$.