1. **Problem statement:** Given that $\overrightarrow{AB} = 2a + 18b$ and point $P$ lies on line $AB$ such that the ratio $AP : PB = 1 : 3$, find the vector $\overrightarrow{AP}$ in terms of $a$ and $b$. 2. **Formula and concept:** When a point $P$ divides a line segment $AB$ in the ratio $m:n$, the position vector of $P$ relative to $A$ is given by: $$\overrightarrow{AP} = \frac{m}{m+n} \overrightarrow{AB}$$ Here, $m=1$ and $n=3$. 3. **Apply the formula:** $$\overrightarrow{AP} = \frac{1}{1+3} \overrightarrow{AB} = \frac{1}{4} (2a + 18b)$$ 4. **Simplify the expression:** $$\overrightarrow{AP} = \frac{1}{4} \times 2a + \frac{1}{4} \times 18b = \frac{2}{4}a + \frac{18}{4}b$$ 5. **Cancel common factors:** $$\overrightarrow{AP} = \cancel{\frac{2}{4}}a + \cancel{\frac{18}{4}}b = \frac{1}{2}a + \frac{9}{2}b$$ 6. **Final answer:** $$\boxed{\overrightarrow{AP} = \frac{1}{2}a + \frac{9}{2}b}$$ This means the vector from $A$ to $P$ is half of $a$ plus $\frac{9}{2}$ times $b$.