If $z$ is a complex number such that $\arg\left(\frac{z-3\sqrt{3}}{z+3\sqrt{3}}\right) = \frac{\pi}{3}$, then the locus of $z$ is:
$|z - 3i| = 6$
$|z - 3i| = 6, \text{Im} z > 0$
$|z - 3i| = 6, \text{Im} z 0)$
Step-by-Step Solution
Key Concept: The number of arithmetic progressions of length 3 depends on whether $n$ is even or odd, determined by counting valid common differences.
Step 1: Analyze the case when $n$ is an even number.
We first consider the scenario where $n$ is an even integer. Let $n = 2m$ for some positive integer $m$. The problem states that the count of triplets with a common difference $d$ is given by $2m - 2d$.
Step 2: Calculate the sum of triplet counts for even $n$.
To find the total number of such triplets for even $n$, we sum the counts for common differences $d$ ranging from $1$ to $m-1$. The sum is:
$$ \sum_{d=1}^{m-1} (2m - 2d) = (2m-2) + (2m-4) + \ldots + (2m-2(m-1)) $$
$$ = (2m-2) + (2m-4) + \ldots + 2 $$
This is an arithmetic progression. We can rewrite it as:
$$ 2(m-1) + 2(m-2) + \ldots + 2(1) $$
Factoring out $2$, we get:
$$ 2 \left( (m-1) + (m-2) + \ldots + 1 \right) $$
The sum of the first $k$ natural numbers is $\frac{k(k+1)}{2}$. Here, $k = m-1$.
So, the sum is:
$$ 2 \left( \frac{(m-1)m}{2} \right) = m(m-1) $$
Now, we express this sum in terms of $n$. Since $n=2m$, we have $m = \frac{n}{2}$. Substituting this into the expression:
$$ m(m-1) = \frac{n}{2} \left( \frac{n}{2} - 1 \right) = \frac{n}{2} \left( \frac{n-2}{2} \right) = \frac{n(n-2)}{4} $$
Step 3: Analyze the case when $n$ is an odd number.
Next, we consider the scenario where $n$ is an odd integer. Let $n = 2m+1$ for some non-negative integer $m$. The problem states that the triplets sum to a series of odd numbers.
$$ 1+3+5+\ldots+(2m-1) $$
This is the sum of the first $m$ odd natural numbers.
Step 4: Calculate the sum of triplet counts for odd $n$.
The sum of the first $k$ odd natural numbers is $k^2$. In this case, the sum is of the first $m$ odd numbers, so the sum is $m^2$.
$$ 1+3+5+\ldots+(2m-1) = m^2 $$
Now, we express this sum in terms of $n$. Since $n=2m+1$, we have $2m = n-1$, which implies $m = \frac{n-1}{2}$. Substituting this into the expression for the sum:
$$ m^2 = \left( \frac{n-1}{2} \right)^2 $$
Step 5: Conclude the derived formulas.
For an even number $n$, the total count of triplets is $\frac{n(n-2)}{4}$.
For an odd number $n$, the total count of triplets is $\left(\frac{n-1}{2}\right)^2$.
Correct Answer: 2,4