Question:

**Step 1:** The sum of the first $k$ terms of an arithmetic progression (A.P.) is given by the formula: $$S_k = \frac{k}{2}[2a + (k-1)d]$$ where $a$ is the first term and $d$ is the common difference. **Step 2:** Use the given condition $S_{2n} = 3S_n$ to establish a relationship between $a$ and $d$. Substitute the sum formula for $S_{2n}$ and $S_n$: $$\frac{2n}{2}[2a + (2n-1)d] = 3 \cdot \frac{n}{2}[2a + (n-1)d]$$ $$n[2a + (2n-1)d] = \frac{3n}{2}[2a + (n-1)d]$$ Assuming $n \neq 0$, divide both sides by $n$: $$2a + (2n-1)d = \frac{3}{2}[2a + (n-1)d]$$ Multiply both sides by 2: $$2(2a + (2n-1)d) = 3(2a + (n-1)d)$$ $$4a + (4n-2)d = 6a + (3n-3)d$$ Rearrange the terms to solve for $a$: $$(4n-2)d - (3n-3)d = 6a - 4a$$ $$(4n-2-3n+3)d = 2a$$ $$(n+1)d = 2a$$ $$a = \frac{(n+1)d}{2}$$ **Step 3:** Calculate $S_{3n}$ by substituting the expression for $a$ into the sum formula: $$S_{3n} = \frac{3n}{2}[2a + (3n-1)d]$$ Substitute $a = \frac{(n+1)d}{2}$: $$S_{3n} = \frac{3n}{2}\left[2\left(\frac{(n+1)d}{2}\right) + (3n-1)d\right]$$ $$S_{3n} = \frac{3n}{2}[(n+1)d + (3n-1)d]$$ $$S_{3n} = \frac{3n}{2}[(n+1) + (3n-1)]d$$ $$S_{3n} = \frac{3n}{2}[4n]d$$ $$S_{3n} = 3n(2n)d$$ $$S_{3n} = 6n^2d$$ **Step 4:** Calculate $S_n$ by substituting the expression for $a$ into the sum formula: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ Substitute $a = \frac{(n+1)d}{2}$: $$S_n = \frac{n}{2}\left[2\left(\frac{(n+1)d}{2}\right) + (n-1)d\right]$$ $$S_n = \frac{n}{2}[(n+1)d + (n-1)d]$$ $$S_n = \frac{n}{2}[(n+1) + (n-1)]d$$ $$S_n = \frac{n}{2}[2n]d$$ $$S_n = n^2d$$ **Step 5:** Determine the ratio $S_{3n}/S_n$: $$\frac{S_{3n}}{S_n} = \frac{6n^2d}{n^2d}$$ Assuming $n \neq 0$ and $d \neq 0$, we can cancel $n^2d$: $$\frac{S_{3n}}{S_n} = 6$$