Question:

**Step 1:** Write the formula for the sum of the first $n$ terms of an arithmetic progression (AP). For an AP with first term $a$ and common difference $d$, the sum of the first $n$ terms is given by: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ **Step 2:** Calculate the sum of the first 5 terms ($S_5$). Using the formula with $n=5$: $$S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}[2a + 4d]$$ $$S_5 = 5(a + 2d) = 5a + 10d$$ **Step 3:** Calculate the sum of the first 10 terms ($S_{10}$). Using the formula with $n=10$: $$S_{10} = \frac{10}{2}[2a + (10-1)d] = 5[2a + 9d]$$ $$S_{10} = 10a + 45d$$ **Step 4:** Apply the given condition that the sum of the first ten terms is four times the sum of the first five terms. $$S_{10} = 4S_5$$ Substitute the expressions for $S_{10}$ and $S_5$: $$10a + 45d = 4(5a + 10d)$$ $$10a + 45d = 20a + 40d$$ **Step 5:** Solve for the ratio of the first term to the common difference, $a:d$. Rearrange the equation to group terms with $a$ and $d$: $$45d - 40d = 20a - 10a$$ $$5d = 10a$$ To find the ratio $a/d$, divide both sides by $10d$: $$\frac{5d}{10d} = \frac{10a}{10d}$$ $$\frac{5}{10} = \frac{a}{d}$$ $$\frac{a}{d} = \frac{1}{2}$$