1. **Problem Statement:** A. Given an arithmetic progression (AP) where the 6th term $a_6=26$ and the 9th term $a_9=17$, find: i. the common difference $d$ ii. the first term $a_1$ iii. the sum of the first 29 terms $S_{29}$ B. Given a geometric progression (GP) where the 6th term $g_6=4$ and the 4th term $g_4=16$, find: i. the common ratio $r$ ii. the first term $g_1$ iii. the sum of the first 5 terms $S_5$ iv. the sum to infinity $S_\infty$ --- ### Part A: Arithmetic Progression 1. The formula for the $n$th term of an AP is: $$a_n = a_1 + (n-1)d$$ 2. Using the given terms: $$a_6 = a_1 + 5d = 26$$ $$a_9 = a_1 + 8d = 17$$ 3. Subtract the first equation from the second to eliminate $a_1$: $$a_9 - a_6 = (a_1 + 8d) - (a_1 + 5d) = 3d = 17 - 26 = -9$$ $$\Rightarrow d = \frac{-9}{3} = -3$$ 4. Substitute $d = -3$ back into $a_6 = a_1 + 5d = 26$: $$a_1 + 5(-3) = 26$$ $$a_1 - 15 = 26$$ $$a_1 = 26 + 15 = 41$$ 5. The sum of the first $n$ terms of an AP is: $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$ 6. Calculate $S_{29}$: $$S_{29} = \frac{29}{2} [2(41) + (29-1)(-3)]$$ $$= \frac{29}{2} [82 + 28(-3)] = \frac{29}{2} [82 - 84] = \frac{29}{2} (-2) = 29 \times (-1) = -29$$ --- ### Part B: Geometric Progression 1. The formula for the $n$th term of a GP is: $$g_n = g_1 r^{n-1}$$ 2. Using the given terms: $$g_6 = g_1 r^5 = 4$$ $$g_4 = g_1 r^3 = 16$$ 3. Divide $g_6$ by $g_4$ to eliminate $g_1$: $$\frac{g_6}{g_4} = \frac{g_1 r^5}{g_1 r^3} = r^2 = \frac{4}{16} = \frac{1}{4}$$ $$\Rightarrow r = \pm \frac{1}{2}$$ 4. Substitute $r = \frac{1}{2}$ into $g_4 = g_1 r^3 = 16$: $$g_1 \left(\frac{1}{2}\right)^3 = 16$$ $$g_1 \times \frac{1}{8} = 16$$ $$g_1 = 16 \times 8 = 128$$ 5. The sum of the first $n$ terms of a GP is: $$S_n = g_1 \frac{1 - r^n}{1 - r}$$ 6. Calculate $S_5$: $$S_5 = 128 \times \frac{1 - (\frac{1}{2})^5}{1 - \frac{1}{2}} = 128 \times \frac{1 - \frac{1}{32}}{\frac{1}{2}} = 128 \times \frac{\frac{31}{32}}{\frac{1}{2}} = 128 \times \frac{31}{32} \times 2$$ $$= 128 \times \frac{62}{32} = 128 \times \frac{31}{16} = 8 \times 31 = 248$$ 7. The sum to infinity of a GP (if $|r|<1$) is: $$S_\infty = \frac{g_1}{1 - r} = \frac{128}{1 - \frac{1}{2}} = \frac{128}{\frac{1}{2}} = 256$$ --- **Final answers:** - Arithmetic progression: - Common difference $d = -3$ - First term $a_1 = 41$ - Sum of first 29 terms $S_{29} = -29$ - Geometric progression: - Common ratio $r = \frac{1}{2}$ - First term $g_1 = 128$ - Sum of first 5 terms $S_5 = 248$ - Sum to infinity $S_\infty = 256$