Let's analyze the problem step by step.

The sequence given is $-9, -6, -3, \ldots$. This is an arithmetic sequence where:

• The first term $a_1 = -9$,
• The common difference $d = -6 - (-9) = 3$.

We need to find the number of terms $n$ such that the sum $S_n$ of the first $n$ terms equals 66.

The sum $S_n$ of the first $n$ terms of an arithmetic sequence can be calculated using the formula:

$S_n = \frac{n}{2} \times \left(2a_1 + (n-1)d\right)$

Substituting the known values: $S_n = \frac{n}{2} \times \left(2(-9) + (n-1) \times 3\right)$ $S_n = \frac{n}{2} \times \left(-18 + 3n - 3\right)$ $S_n = \frac{n}{2} \times (3n - 21)$ $S_n = \frac{n}{2} \times 3(n - 7)$ $S_n = \frac{3n(n - 7)}{2}$

We need to find $n$ such that $S_n = 66$:

$\frac{3n(n - 7)}{2} = 66$

Multiply both sides by 2:

$3n(n - 7) = 132$

Divide both sides by 3:

$n(n - 7) = 44$

Now, solve the quadratic equation:

$n^2 - 7n - 44 = 0$

Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -7$, and $c = -44$:

$n = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-44)}}{2(1)}$ $n = \frac{7 \pm \sqrt{49 + 176}}{2}$ $n = \frac{7 \pm \sqrt{225}}{2}$ $n = \frac{7 \pm 15}{2}$

This gives two possible solutions:

$n = \frac{22}{2} = 11 \quad \text{or} \quad n = \frac{-8}{2} = -4$

Since $n$ must be a positive integer, $n = 11$.

Conclusion: The sum of the first 11 terms of the sequence is 66.

Would you like further details or have any questions? Here are some related questions:

1. How do we derive the sum formula for an arithmetic sequence?
2. What happens if the common difference is negative?
3. Can this method be applied to geometric sequences as well?
4. How does the quadratic formula work in solving equations?
5. What if the sequence had a fractional common difference?

Tip: When working with arithmetic sequences, identifying the first term and common difference is crucial to solving problems related to sums and specific terms.