1. **Problem statement:** We have two sequences: an arithmetic progression $\{a_n\}$ and a geometric progression $\{b_n\}$. Given: - $b_1 + b_2 + b_3 = 65$ - $b_6 > b_5$ - $b_1 - a_1 = 1$ - $b_2 - a_2 = 8$ - $b_3 - a_3 = 35$ - $b_1 + b_2 + \cdots + b_n = 200$ Find $n$. 2. **Formulas and definitions:** - Arithmetic progression: $a_n = a_1 + (n-1)d$ where $d$ is the common difference. - Geometric progression: $b_n = b_1 r^{n-1}$ where $r$ is the common ratio. - Sum of first $n$ terms of geometric progression: $$S_n = b_1 \frac{r^n - 1}{r - 1}$$ 3. **Use given conditions:** From $b_1 + b_2 + b_3 = 65$: $$b_1 + b_1 r + b_1 r^2 = b_1(1 + r + r^2) = 65$$ From $b_6 > b_5$: $$b_1 r^5 > b_1 r^4 \implies r > 1$$ From differences: $$b_1 - a_1 = 1$$ $$b_1 r - (a_1 + d) = 8$$ $$b_1 r^2 - (a_1 + 2d) = 35$$ 4. **Express $a_1$ from first difference:** $$a_1 = b_1 - 1$$ Substitute into second difference: $$b_1 r - (b_1 - 1 + d) = 8 \implies b_1 r - b_1 + 1 - d = 8$$ $$\Rightarrow d = b_1 r - b_1 + 1 - 8 = b_1 (r - 1) - 7$$ Substitute into third difference: $$b_1 r^2 - (b_1 - 1 + 2d) = 35$$ $$b_1 r^2 - b_1 + 1 - 2d = 35$$ $$-2d = 35 - b_1 r^2 + b_1 - 1$$ $$d = \frac{b_1 r^2 - b_1 + 1 - 35}{2} = \frac{b_1 (r^2 - 1) - 34}{2}$$ 5. **Equate the two expressions for $d$:** $$b_1 (r - 1) - 7 = \frac{b_1 (r^2 - 1) - 34}{2}$$ Multiply both sides by 2: $$2 b_1 (r - 1) - 14 = b_1 (r^2 - 1) - 34$$ Bring all terms to one side: $$2 b_1 (r - 1) - b_1 (r^2 - 1) = -34 + 14 = -20$$ Expand: $$2 b_1 r - 2 b_1 - b_1 r^2 + b_1 = -20$$ $$b_1 (2 r - 2 - r^2 + 1) = -20$$ $$b_1 (-r^2 + 2 r - 1) = -20$$ Note that $-r^2 + 2 r - 1 = -(r^2 - 2 r + 1) = -(r - 1)^2$ So: $$b_1 (-(r - 1)^2) = -20 \implies -b_1 (r - 1)^2 = -20$$ Multiply both sides by -1: $$b_1 (r - 1)^2 = 20$$ 6. **Recall from step 3:** $$b_1 (1 + r + r^2) = 65$$ 7. **Use substitution:** Let $x = r - 1$, then $r = x + 1$. Calculate $1 + r + r^2$ in terms of $x$: $$1 + (x + 1) + (x + 1)^2 = 1 + x + 1 + x^2 + 2x + 1 = x^2 + 3x + 3$$ From step 5: $$b_1 x^2 = 20$$ From step 6: $$b_1 (x^2 + 3x + 3) = 65$$ 8. **Express $b_1$ from step 5:** $$b_1 = \frac{20}{x^2}$$ Substitute into step 6: $$\frac{20}{x^2} (x^2 + 3x + 3) = 65$$ Multiply both sides by $x^2$: $$20 (x^2 + 3x + 3) = 65 x^2$$ Expand left side: $$20 x^2 + 60 x + 60 = 65 x^2$$ Bring all terms to one side: $$20 x^2 + 60 x + 60 - 65 x^2 = 0$$ $$-45 x^2 + 60 x + 60 = 0$$ Divide entire equation by 15: $$-3 x^2 + 4 x + 4 = 0$$ Multiply both sides by -1: $$3 x^2 - 4 x - 4 = 0$$ 9. **Solve quadratic equation:** $$3 x^2 - 4 x - 4 = 0$$ Discriminant: $$\Delta = (-4)^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64$$ Roots: $$x = \frac{4 \pm \sqrt{64}}{2 \cdot 3} = \frac{4 \pm 8}{6}$$ Two solutions: - $$x_1 = \frac{4 + 8}{6} = 2$$ - $$x_2 = \frac{4 - 8}{6} = -\frac{2}{3}$$ 10. **Recall $r = x + 1$ and $r > 1$:** - For $x_1 = 2$, $r = 3 > 1$ (valid) - For $x_2 = -\frac{2}{3}$, $r = \frac{1}{3} < 1$ (not valid) So, $r = 3$. 11. **Find $b_1$:** $$b_1 = \frac{20}{x^2} = \frac{20}{2^2} = \frac{20}{4} = 5$$ 12. **Sum of first $n$ terms of geometric progression:** $$S_n = b_1 \frac{r^n - 1}{r - 1} = 200$$ Substitute $b_1 = 5$, $r = 3$: $$5 \frac{3^n - 1}{3 - 1} = 200$$ $$5 \frac{3^n - 1}{2} = 200$$ Multiply both sides by 2: $$5 (3^n - 1) = 400$$ Divide both sides by 5: $$3^n - 1 = 80$$ $$3^n = 81$$ 13. **Solve for $n$:** $$3^n = 3^4$$ $$\Rightarrow n = 4$$ **Final answer:** $$\boxed{4}$$