1. Muammo: Berilgan arifmetik progressiya hadlari $\{a_n\}$ va geometrik progressiya hadlari $\{b_n\}$ uchun shartlar berilgan: $b_1 + b_2 + b_3 = 65$, $b_6 > b_5$, $b_1 - a_1 = 1$, $b_2 - a_2 = 8$, $b_3 - a_3 = 35$, va $b_1 + b_2 + \cdots + b_n = 200$. $n$ ni topish kerak. 2. Formulalar va qoidalar: - Arifmetik progressiya hadlari: $a_n = a_1 + (n-1)d$. - Geometrik progressiya hadlari: $b_n = b_1 r^{n-1}$. - Geometrik progressiya yig'indisi: $S_n = b_1 \frac{r^n - 1}{r - 1}$, agar $r \neq 1$. 3. Berilgan shartlarni yozamiz: - $b_1 + b_2 + b_3 = b_1 + b_1 r + b_1 r^2 = b_1(1 + r + r^2) = 65$. - $b_6 > b_5$ degani $b_1 r^5 > b_1 r^4$ yoki $r > 1$ (chunki $b_1 > 0$). - $b_1 - a_1 = 1$. - $b_2 - a_2 = b_1 r - (a_1 + d) = 8$. - $b_3 - a_3 = b_1 r^2 - (a_1 + 2d) = 35$. - $b_1 + b_2 + \cdots + b_n = b_1 \frac{r^n - 1}{r - 1} = 200$. 4. $a_1$ ni ifodalash uchun birinchi tenglamadan: $$a_1 = b_1 - 1$$ 5. Ikkinchi tenglamadan: $$b_1 r - (a_1 + d) = 8 \Rightarrow b_1 r - (b_1 - 1 + d) = 8 \Rightarrow b_1 r - b_1 + 1 - d = 8$$ $$\Rightarrow b_1 (r - 1) - d = 7 \Rightarrow d = b_1 (r - 1) - 7$$ 6. Uchinchi tenglamadan: $$b_1 r^2 - (a_1 + 2d) = 35 \Rightarrow b_1 r^2 - (b_1 - 1 + 2d) = 35$$ $$\Rightarrow b_1 r^2 - b_1 + 1 - 2d = 35 \Rightarrow b_1 (r^2 - 1) - 2d = 34$$ 7. $d$ ni 5-qadamdagi ifodadan o'rniga qo'yamiz: $$b_1 (r^2 - 1) - 2[b_1 (r - 1) - 7] = 34$$ $$b_1 (r^2 - 1) - 2 b_1 (r - 1) + 14 = 34$$ $$b_1 (r^2 - 1 - 2r + 2) = 20$$ $$b_1 (r^2 - 2r + 1) = 20$$ $$b_1 (r - 1)^2 = 20$$ 8. Endi birinchi tenglamadan $b_1 (1 + r + r^2) = 65$ va yuqoridagi tenglamadan $b_1 (r - 1)^2 = 20$. 9. $b_1$ ni ifodalab, tenglamalarni taqqoslaymiz: $$b_1 = \frac{65}{1 + r + r^2} = \frac{20}{(r - 1)^2}$$ 10. Tenglama: $$\frac{65}{1 + r + r^2} = \frac{20}{(r - 1)^2} \Rightarrow 65 (r - 1)^2 = 20 (1 + r + r^2)$$ 11. Kengaytiramiz: $$65 (r^2 - 2r + 1) = 20 + 20 r + 20 r^2$$ $$65 r^2 - 130 r + 65 = 20 + 20 r + 20 r^2$$ 12. Hammasini chapga o'tkazamiz: $$65 r^2 - 130 r + 65 - 20 - 20 r - 20 r^2 = 0$$ $$45 r^2 - 150 r + 45 = 0$$ 13. 15 ga bo'lamiz: $$3 r^2 - 10 r + 3 = 0$$ 14. Kvadrat tenglamani yechamiz: $$r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}$$ 15. Ikkita ildiz: $$r_1 = \frac{10 + 8}{6} = 3$$ $$r_2 = \frac{10 - 8}{6} = \frac{1}{3}$$ 16. $r > 1$ shartiga ko'ra $r = 3$ ni olamiz. 17. $b_1$ ni topamiz: $$b_1 = \frac{65}{1 + 3 + 9} = \frac{65}{13} = 5$$ 18. $a_1$ ni topamiz: $$a_1 = b_1 - 1 = 5 - 1 = 4$$ 19. $d$ ni topamiz: $$d = b_1 (r - 1) - 7 = 5 (3 - 1) - 7 = 10 - 7 = 3$$ 20. Endi $b_1 + b_2 + \cdots + b_n = 200$ tenglamani yechamiz: $$5 \frac{3^n - 1}{3 - 1} = 200 \Rightarrow 5 \frac{3^n - 1}{2} = 200$$ 21. Ko'paytmani yechamiz: $$\frac{5}{2} (3^n - 1) = 200 \Rightarrow 3^n - 1 = \frac{200 \cdot 2}{5} = 80$$ 22. $3^n = 81$ bo'ladi. 23. $3^n = 3^4$ ekan, shuning uchun: $$n = 4$$ Javob: $n = 4$