For what values of t does the geometric sequence T1 = -t2 - 6t - 9, T2 = (t3 + 9t2 + 27t + 27)/2, ... converge?

The first term is $T_1 = -t^2 - 6t - 9$.
We can factor out $-1$ and recognize a perfect square trinomial:

$$T_1 = -(t^2 + 6t + 9) = -(t+3)^2$$

The second term is $T_2 = \frac{t^3 + 9t^2 + 27t + 27}{2}$.
The numerator is a perfect cube expansion, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, where $a=t$ and $b=3$:

$$T_2 = \frac{(t+3)^3}{2}$$

The common ratio $r$ is given by $r = \frac{T_2}{T_1}$.
We must consider the case where $T_1 = 0$. If $T_1 = -(t+3)^2 = 0$, then $t+3=0$, so $t=-3$.
If $t=-3$, then $T_1 = 0$ and $T_2 = \frac{(-3+3)^3}{2} = 0$. The sequence is $0, 0, 0, ...$, which converges. So $t=-3$ is a value for which the sequence converges.

Now, for $t \neq -3$, $T_1 \neq 0$, and we can calculate $r$:

$$r = \frac{\frac{(t+3)^3}{2}}{-(t+3)^2}$$ $$r = \frac{(t+3)^3}{2 \cdot (-(t+3)^2)}$$ $$r = -\frac{t+3}{2}$$

A geometric sequence converges if its common ratio $r$ satisfies $|r| < 1$.
We need to solve the inequality:

$$\left|-\frac{t+3}{2}\right| < 1$$

This simplifies to:

$$\frac{|t+3|}{2} < 1$$

Multiply both sides by 2:

$$|t+3| < 2$$

The inequality $|t+3| < 2$ can be written as:

$$-2 < t+3 < 2$$

Subtract 3 from all parts of the inequality:

$$-2 - 3 < t < 2 - 3$$ $$-5 < t < -1$$

The interval $-5 < t < -1$ includes the value $t=-3$. When $t=-3$, $r = -\frac{-3+3}{2} = 0$, and $|0|<1$, so the condition $|r|<1$ is satisfied. Therefore, the sequence converges for all $t$ in the interval.

The values of $t$ for which the sequence will converge are:

$$\boxed{-5 < t < -1}$$