1. The first problem asks for the sum of the first four terms of a geometric sequence where the first term $G_1 = \frac{1}{4}$ and the common ratio $r = 3$. 2. The formula for the sum of the first $n$ terms of a geometric sequence is: $$S_n = G_1 \frac{r^n - 1}{r - 1}$$ 3. Substitute $G_1 = \frac{1}{4}$, $r = 3$, and $n = 4$: $$S_4 = \frac{1}{4} \times \frac{3^4 - 1}{3 - 1} = \frac{1}{4} \times \frac{81 - 1}{2} = \frac{1}{4} \times \frac{80}{2} = \frac{1}{4} \times 40 = 10$$ 4. So, the sum of the first four terms is $10$. 5. The second problem asks for the sum of the first $n$ terms of the arithmetic sequence defined by $S_n = \sum_{k=1}^n (5k - 3)$. 6. To find the sum, we use the formula for the sum of an arithmetic series: $$S_n = \frac{n}{2} (a_1 + a_n)$$ where $a_1$ is the first term and $a_n$ is the $n$th term. 7. The first term $a_1 = 5(1) - 3 = 2$. 8. The $n$th term $a_n = 5n - 3$. 9. Substitute into the sum formula: $$S_n = \frac{n}{2} (2 + (5n - 3)) = \frac{n}{2} (5n - 1) = \frac{5n^2 - n}{2}$$ 10. This matches the given formula option $\frac{6n^2 - n}{2}$ only if the coefficient of $n^2$ is 6, but here it is 5, so the correct sum is $\frac{5n^2 - n}{2}$. 11. Therefore, the sum of the first $n$ terms is $\frac{5n^2 - n}{2}$, which is not exactly any of the options given, but closest to option 2 if the problem intended $5n^2$ instead of $6n^2$. Final answers: - Sum of first four terms of geometric sequence: $10$ - Sum of first $n$ terms of arithmetic sequence: $\frac{5n^2 - n}{2}$