Let the first term of the geometric series be $a$ and the common ratio be $r$.
The formula for the $n$-th term of a geometric series is given by:

$T_n = a r^{n-1}$

Given:

• Fourth term: $T_4 = a r^{4-1} = a r^3 = 30$
• Ninth term: $T_9 = a r^{9-1} = a r^8 = 960$

Step 1: Form equations from the given terms

a r^3 = 30 \tag{1} a r^8 = 960 \tag{2}

Step 2: Divide equation (2) by equation (1) to eliminate $a$

$\frac{a r^8}{a r^3} = \frac{960}{30}$ $r^5 = 32$

Since $32 = 2^5$, we get $r = 2$

Step 3: Substitute $r = 2$ back into equation (1) to find $a$

$a (2)^3 = 30$ $a \cdot 8 = 30 \implies a = \frac{30}{8} = \frac{15}{4}$

Step 4: Calculate the sum of the first nine terms

The formula for the sum of the first $n$ terms of a geometric series is:

$S_n = a \frac{r^n - 1}{r - 1}$

Substitute $a = \frac{15}{4}$, $r = 2$, and $n = 9$:

$S_9 = \frac{15}{4} \cdot \frac{2^9 - 1}{2 - 1}$

$= \frac{15}{4} \cdot (512 - 1) = \frac{15}{4} \cdot 511$

$= \frac{7665}{4} = 1916.25$

Final Answer:

$\boxed{1916.25}$

Do you want details or have any questions?

Here are 5 related questions:

1. How would the solution change if the ninth term was different?
2. What if the common ratio were negative? How would the series behave?
3. How does the sum formula adapt for an infinite geometric series?
4. Can a geometric series ever converge with a common ratio of 2?
5. Why is the first term $a$ crucial in determining the entire sequence?

Tip: Remember, a geometric series converges only if the common ratio's absolute value is less than 1!