1. **Problem statement:** The first, fifth, and eighth terms of an arithmetic progression (AP) with common difference $3$ form the first three terms of a geometric progression (GP). We need to find the first term and common ratio of the GP, and the sum of the first 9 terms of the AP. 2. **Formulas and rules:** - The $n$th term of an AP is given by $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. - The terms of the GP satisfy $$t_2^2 = t_1 \times t_3$$ where $t_1, t_2, t_3$ are the first three terms. - The sum of the first $n$ terms of an AP is $$S_n = \frac{n}{2}[2a + (n-1)d]$$. 3. **Express the terms:** - First term of AP: $a_1 = a$ - Fifth term of AP: $a_5 = a + 4d = a + 4 \times 3 = a + 12$ - Eighth term of AP: $a_8 = a + 7d = a + 7 \times 3 = a + 21$ 4. **Apply GP condition:** The terms $a_1$, $a_5$, $a_8$ form a GP, so $$ (a + 12)^2 = a \times (a + 21) $$ 5. **Solve the quadratic:** $$ (a + 12)^2 = a(a + 21) $$ $$ a^2 + 24a + 144 = a^2 + 21a $$ Subtract $a^2$ from both sides: $$ 24a + 144 = 21a $$ $$ 24a - 21a = -144 $$ $$ 3a = -144 $$ $$ a = -48 $$ 6. **Find the common ratio $r$ of the GP:** $$ r = \frac{a_5}{a_1} = \frac{a + 12}{a} = \frac{-48 + 12}{-48} = \frac{-36}{-48} = \frac{3}{4} $$ 7. **Sum of first 9 terms of AP:** $$ S_9 = \frac{9}{2}[2a + (9-1)d] = \frac{9}{2}[2(-48) + 8 \times 3] = \frac{9}{2}[-96 + 24] = \frac{9}{2}[-72] = 9 \times (-36) = -324 $$ **Final answers:** - First term of AP: $-48$ - Common ratio of GP: $\frac{3}{4}$ - Sum of first 9 terms of AP: $-324$