1. **Problem statement:** We have an arithmetic progression (AP) with first term $a_1=8$ and common difference $d \neq 0$. The 1st, 5th, and 8th terms of this AP form the 1st, 2nd, and 3rd terms of a geometric progression (GP) with common ratio $r$. 2. **Formulating the terms:** - AP terms: $a_1=8$, $a_5=8+4d$, $a_8=8+7d$ - GP terms: $g_1=8$, $g_2=8+4d$, $g_3=8+7d$ 3. **GP property:** The ratio between consecutive terms is constant: $$r=\frac{g_2}{g_1}=\frac{8+4d}{8}$$ $$r=\frac{g_3}{g_2}=\frac{8+7d}{8+4d}$$ 4. **Equations connecting $d$ and $r$:** From the above, $$r=\frac{8+4d}{8}$$ $$r=\frac{8+7d}{8+4d}$$ 5. **Equate the two expressions for $r$:** $$\frac{8+4d}{8} = \frac{8+7d}{8+4d}$$ Cross-multiply: $$(8+4d)(8+4d) = 8(8+7d)$$ Expand: $$64 + 64d + 16d^2 = 64 + 56d$$ Simplify: $$16d^2 + 64d - 56d = 64 - 64$$ $$16d^2 + 8d = 0$$ 6. **Solve for $d$:** Factor out $8d$: $$8d(2d + 1) = 0$$ Since $d \neq 0$, $$2d + 1 = 0 \implies d = -\frac{1}{2}$$ 7. **Find $r$ using $r=\frac{8+4d}{8}$:** $$r = \frac{8 + 4(-\frac{1}{2})}{8} = \frac{8 - 2}{8} = \frac{6}{8} = \frac{3}{4}$$ 8. **Sum to infinity of the GP:** Since $|r| = \frac{3}{4} < 1$, the sum to infinity exists and is given by: $$S_\infty = \frac{g_1}{1-r} = \frac{8}{1 - \frac{3}{4}} = \frac{8}{\frac{1}{4}} = 32$$ 9. **Sum of first 8 terms of the AP:** The sum of first $n$ terms of an AP is: $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$ For $n=8$: $$S_8 = \frac{8}{2} [2(8) + 7(-\frac{1}{2})] = 4 [16 - \frac{7}{2}] = 4 \times \frac{32 - 7}{2} = 4 \times \frac{25}{2} = 50$$ **Final answers:** - $r = \frac{3}{4}$ - $d = -\frac{1}{2}$ - Sum to infinity of GP = 32 - Sum of first 8 terms of AP = 50