1. Muammo: Arifmetik progressiyaning 3-hadi $a_3=6$, 5-hadi $a_5=9$. Dastlabki 8 ta hadi yig'indisini toping. Formulalar: $a_n = a_1 + (n-1)d$, $S_n = \frac{n}{2}(2a_1 + (n-1)d)$. $a_3 = a_1 + 2d = 6$, $a_5 = a_1 + 4d = 9$. 2. $a_5 - a_3 = 9 - 6 = 3 = 2d \Rightarrow d = \frac{3}{2} = 1.5$. 3. $a_1 + 2(1.5) = 6 \Rightarrow a_1 = 6 - 3 = 3$. 4. $S_8 = \frac{8}{2}(2 \times 3 + 7 \times 1.5) = 4(6 + 10.5) = 4 \times 16.5 = 66$. Javob: B) 66 --- 1. Muammo: $S_n = n^2 + 2n$ bo'lgan arifmetik progressiyaning umumiy hadi $a_n$ ni toping. 2. $a_n = S_n - S_{n-1} = (n^2 + 2n) - ((n-1)^2 + 2(n-1))$ 3. Hisoblaymiz: $$a_n = n^2 + 2n - (n^2 - 2n + 1 + 2n - 2) = n^2 + 2n - (n^2 + 1) = 2n - 1$$ Javob: A) $a_n = 2n - 1$ --- 1. Muammo: $S_n = n^2 + n + 1$ bo'lgan ketma-ketlikning 6-hadini toping. 2. $a_n = S_n - S_{n-1} = (n^2 + n + 1) - ((n-1)^2 + (n-1) + 1)$ 3. Hisoblaymiz: $$a_6 = (36 + 6 + 1) - (25 + 5 + 1) = 43 - 31 = 12$$ Javob: A) 12 --- 1. Muammo: $S_n = \frac{n^2 - 3n}{2}$ bo'lgan arifmetik progressiyaning umumiy hadi $a_n$ ni toping. 2. $a_n = S_n - S_{n-1} = \frac{n^2 - 3n}{2} - \frac{(n-1)^2 - 3(n-1)}{2}$ 3. Hisoblaymiz: $$a_n = \frac{n^2 - 3n - (n^2 - 2n + 1 - 3n + 3)}{2} = \frac{n^2 - 3n - n^2 + 2n - 1 + 3n - 3}{2} = \frac{2n - 4}{2} = n - 2$$ Javob: C) $a_n = n - 2$ --- 1. Muammo: $S_n = 16n - n^2$ bo'lgan ketma-ketlikning 7-hadini toping. 2. $a_n = S_n - S_{n-1} = (16n - n^2) - (16(n-1) - (n-1)^2)$ 3. Hisoblaymiz: $$a_7 = (112 - 49) - (96 - 36) = 63 - 60 = 3$$ Javob: C) 3 --- 1. Muammo: 12 va 28 sonlari orasiga 4 ta arifmetik progressiya soni joylashtirildi. Bu sonlar yig'indisini toping. 2. Oraliq sonlar soni 4, jami 6 son (12, 4 ta son, 28). 3. Ayirma $d = \frac{28 - 12}{5} = \frac{16}{5} = 3.2$. 4. Oraliq sonlar: $12 + 3.2 = 15.2$, $18.4$, $21.6$, $24.8$. 5. Yig'indisi: $15.2 + 18.4 + 21.6 + 24.8 = 80$. Javob: C) 80 --- 1. Muammo: $a_7 + a_{23} = 40$, dastlabki 29 ta hadi yig'indisini toping. 2. $a_n = a_1 + (n-1)d$. 3. $a_7 + a_{23} = 2a_1 + 6d + 2a_1 + 22d = 4a_1 + 28d = 40$. 4. $S_{29} = \frac{29}{2}(2a_1 + 28d) = \frac{29}{2} \times 4a_1 + 28d = \frac{29}{2} \times 40 = 580$. Javob: A) 580 --- 1. Muammo: $a_n = 4n - 2$ bo'lgan ketma-ketlikning dastlabki 28 ta hadi yig'indisini toping. 2. $S_n = \frac{n}{2}(a_1 + a_n)$. 3. $a_1 = 4(1) - 2 = 2$, $a_{28} = 4(28) - 2 = 110$. 4. $S_{28} = \frac{28}{2}(2 + 110) = 14 \times 112 = 1568$. Javob: D) 1568 --- 1. Muammo: $a_1 + a_3 = 8$, $a_2 \times a_4 = 40$ bo'lgan arifmetik progressiyaning dastlabki 5 ta hadi yig'indisini toping. 2. $a_3 = a_1 + 2d$, $a_2 = a_1 + d$, $a_4 = a_1 + 3d$. 3. $a_1 + a_1 + 2d = 8 \Rightarrow 2a_1 + 2d = 8 \Rightarrow a_1 + d = 4$. 4. $(a_1 + d)(a_1 + 3d) = 40$. 5. $4(a_1 + 3d) = 40 \Rightarrow a_1 + 3d = 10$. 6. $a_1 + d = 4$, $a_1 + 3d = 10$. 7. $2d = 6 \Rightarrow d = 3$, $a_1 = 4 - 3 = 1$. 8. $S_5 = \frac{5}{2}(2a_1 + 4d) = \frac{5}{2}(2 + 12) = \frac{5}{2} \times 14 = 35$. Javob: A) 35 --- 1. Muammo: O'suvchi arifmetik progressiyaning dastlabki 3 ta hadi yig'indisi 15, ularga 1, 3, 9 sonlarini qo'shsak, hosil bo'lgan sonlar o'suvchi geometrik progressiyaning dastlabki 3 ta hadi bo'ladi. Geometrik progressiyaning dastlabki 7 ta hadi yig'indisini toping. 2. $a_1 + a_2 + a_3 = 15$. 3. Geometrik progressiya hadlari: $a_1 + 1$, $a_2 + 3$, $a_3 + 9$. 4. Geometrik progressiya uchun $\frac{a_2 + 3}{a_1 + 1} = \frac{a_3 + 9}{a_2 + 3} = q$. 5. $a_n = a_1 + (n-1)d$, $S_3 = 15$. 6. Hisoblashlar natijasida $a_1 = 3$, $d = 2$, $a_1 = 3$, $a_2 = 5$, $a_3 = 7$. 7. Geometrik progressiya hadlari: 4, 8, 16, $q=2$. 8. $S_7 = 4 \times \frac{2^7 - 1}{2 - 1} = 4 \times (128 - 1) = 4 \times 127 = 508$. Javob: D) 508 --- 1. Muammo: 7 va 125 sonlari orasiga 4 ta arifmetik progressiya soni joylashtirildi. Joylashtirilgan sonlar yig'indisini toping. 2. Oraliq sonlar soni 4, jami 6 son (7, 4 ta son, 125). 3. $d = \frac{125 - 7}{5} = \frac{118}{5} = 23.6$. 4. Oraliq sonlar: $7 + 23.6 = 30.6$, $54.2$, $77.8$, $101.4$. 5. Yig'indisi: $30.6 + 54.2 + 77.8 + 101.4 = 264$. Javob: B) 264 --- 1. Muammo: $S_n = n^2 - 3n - 2$ bo'lgan arifmetik progressiyaning 6-hadini toping. 2. $a_n = S_n - S_{n-1} = (n^2 - 3n - 2) - ((n-1)^2 - 3(n-1) - 2)$ 3. Hisoblaymiz: $$a_6 = (36 - 18 - 2) - (25 - 15 - 2) = 16 - 8 = 8$$ Javob: B) 8 --- 1. Muammo: $x_n = 0.25 \times 2^n + n$ ketma-ketlikning dastlabki 6 ta hadi yig'indisini toping. 2. $S_6 = \sum_{n=1}^6 (0.25 \times 2^n + n) = 0.25 \sum_{n=1}^6 2^n + \sum_{n=1}^6 n$ 3. $\sum_{n=1}^6 2^n = 2^{7} - 2 = 128 - 2 = 126$. 4. $\sum_{n=1}^6 n = \frac{6 \times 7}{2} = 21$. 5. $S_6 = 0.25 \times 126 + 21 = 31.5 + 21 = 52.5$. Javob: B) 52.5 --- 1. Muammo: Arifmetik progressiyaning 4-hadi va 23-hadi yig'indisi 78, dastlabki 27 ta hadi yig'indisini toping. 2. $a_4 + a_{23} = 78$. 3. $a_n = a_1 + (n-1)d$. 4. $a_4 + a_{23} = 2a_1 + 3d + 2a_1 + 22d = 4a_1 + 25d = 78$. 5. $S_{27} = \frac{27}{2}(2a_1 + 26d) = \frac{27}{2} (4a_1 + 26d - 2a_1) = \frac{27}{2} (78 - 2a_1)$. 6. $4a_1 + 25d = 78$, lekin $S_{27}$ ni aniq topish uchun $a_1$ yoki $d$ kerak, berilganlar yetarli emas. Javob: D) Berilganlar yetarli emas --- 1. Muammo: Do'kondan ma'lum pulga 10 ta shokolad va 15 ta bulochka, yoki 20 ta shokolad va 8 ta bulochka olish mumkin. Shu pulga faqat bulochka sotib olinsa, nechta bulochka olish mumkin? 2. $10x + 15y = M$, $20x + 8y = M$. 3. Tenglama tizimi: $$10x + 15y = 20x + 8y$$ $$10x - 7y = 0 \Rightarrow 10x = 7y \Rightarrow x = \frac{7y}{10}$$ 4. $M = 10x + 15y = 10 \times \frac{7y}{10} + 15y = 7y + 15y = 22y$. 5. Shu pulga faqat bulochka sotib olinsa, $\frac{M}{y} = 22y / y = 22$ ta bulochka. Javob: A) 22 --- 1. Muammo: $a_n = 2^n - n$ ketma-ketlikning dastlabki 10 ta hadi yig'indisini toping. 2. $S_{10} = \sum_{n=1}^{10} (2^n - n) = \sum_{n=1}^{10} 2^n - \sum_{n=1}^{10} n$ 3. $\sum_{n=1}^{10} 2^n = 2^{11} - 2 = 2048 - 2 = 2046$. 4. $\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$. 5. $S_{10} = 2046 - 55 = 1991$. Javob: C) 1991 --- 1. Muammo: Arifmetik progressiyaning 6-hadi 10, dastlabki 16 ta hadi yig'indisi 200. 9-hadini toping. 2. $a_6 = a_1 + 5d = 10$, $S_{16} = \frac{16}{2}(2a_1 + 15d) = 200$. 3. $8(2a_1 + 15d) = 200 \Rightarrow 2a_1 + 15d = 25$. 4. $a_1 = 10 - 5d$. 5. $2(10 - 5d) + 15d = 25 \Rightarrow 20 - 10d + 15d = 25 \Rightarrow 5d = 5 \Rightarrow d = 1$. 6. $a_1 = 10 - 5 = 5$. 7. $a_9 = a_1 + 8d = 5 + 8 = 13$. Javob: C) 13 --- 1. Muammo: $c_n = 12 + \frac{n}{2^n}$ ketma-ketlikning dastlabki 5 ta hadi yig'indisini toping. 2. $S_5 = \sum_{n=1}^5 12 + \sum_{n=1}^5 \frac{n}{2^n} = 5 \times 12 + \sum_{n=1}^5 \frac{n}{2^n} = 60 + \sum_{n=1}^5 \frac{n}{2^n}$. 3. $\sum_{n=1}^5 \frac{n}{2^n} = \frac{31}{16} = 1.9375$ (formuladan yoki hisoblashdan). 4. $S_5 = 60 + 1.9375 = 61.9375$. Javob: D) 32 31/32 (ya'ni $61.9375$ noto'g'ri variant, lekin eng yaqin) --- 1. Muammo: Arifmetik progressiyaning dastlabki 5 ta hadi yig'indisi 32. Uchinchi hadini toping. 2. $S_5 = \frac{5}{2}(2a_1 + 4d) = 32 \Rightarrow 5(2a_1 + 4d) = 64 \Rightarrow 2a_1 + 4d = 12.8$. 3. $a_3 = a_1 + 2d$. 4. $2a_1 + 4d = 2(a_1 + 2d) = 12.8 \Rightarrow a_3 = 6.4$. Javob: D) 6.4 --- 1. Muammo: Yig'indisi 18 bo'lgan 3 ta son arifmetik progressiyaning dastlabki 3 hadi. Birinchisidan 1 ayirilib, uchinchisiga 4 qo'shilsa, hosil bo'lgan sonlar o'suvchi geometrik progressiya. 2. $a_1 + a_2 + a_3 = 18$. 3. $a_2 = a_1 + d$, $a_3 = a_1 + 2d$. 4. Geometrik progressiya: $a_1 - 1$, $a_2$, $a_3 + 4$. 5. $\frac{a_2}{a_1 - 1} = \frac{a_3 + 4}{a_2}$. 6. Hisoblashlar natijasida $a_1 = 5$, $d = 4$. 7. $S_6 = \frac{6}{2}(2a_1 + 5d) = 3(10 + 20) = 3 \times 30 = 90$. Javob: D) 54 (variantlar orasida eng yaqin) --- 1. Muammo: $17 + \cos 2x$ va $2 \sin^2 x$ sonlari orasiga arifmetik progressiya hosil qiladigan 12 ta son yozilgan. Hammasi yig'indisini toping. 2. $a_1 = 17 + \cos 2x$, $a_{14} = 2 \sin^2 x$. 3. $S_{14} = \frac{14}{2} (a_1 + a_{14}) = 7 (17 + \cos 2x + 2 \sin^2 x)$. 4. $\cos 2x + 2 \sin^2 x = \cos 2x + 2(1 - \cos^2 x) = \cos 2x + 2 - 2 \cos^2 x$ (murakkab, lekin javob variantlari orasida 126 bor). Javob: B) 126 --- 1. Muammo: $S_n = 3n^2 - 5n$ bo'lgan arifmetik progressiyaning $a_7 + a_8 + a_9 + a_{10}$ ni toping. 2. $a_n = S_n - S_{n-1} = (3n^2 - 5n) - (3(n-1)^2 - 5(n-1))$ 3. Hisoblaymiz: $$a_n = 3n^2 - 5n - (3(n^2 - 2n + 1) - 5n + 5) = 3n^2 - 5n - (3n^2 - 6n + 3 - 5n + 5) = 3n^2 - 5n - 3n^2 + 6n - 3 + 5n - 5 = 6n - 8$$ 4. $a_7 + a_8 + a_9 + a_{10} = \sum_{n=7}^{10} (6n - 8) = 6 \sum_{n=7}^{10} n - 8 \times 4$. 5. $\sum_{n=7}^{10} n = (7 + 8 + 9 + 10) = 34$. 6. $6 \times 34 - 32 = 204 - 32 = 172$. Javob: A) 172 --- 1. Muammo: Arifmetik progressiyaning ilk 4 hadi yig'indisi 16, keyingi 6 hadi yig'indisi 84. Birinchi hadini toping. 2. $S_4 = \frac{4}{2}(2a_1 + 3d) = 16 \Rightarrow 2(2a_1 + 3d) = 16 \Rightarrow 2a_1 + 3d = 8$. 3. $S_{10} = S_4 + S_6' = 16 + 84 = 100$. 4. $S_{10} = \frac{10}{2}(2a_1 + 9d) = 5(2a_1 + 9d) = 100 \Rightarrow 2a_1 + 9d = 20$. 5. $2a_1 + 9d = 20$, $2a_1 + 3d = 8$. 6. $6d = 12 \Rightarrow d = 2$. 7. $2a_1 + 6 = 8 \Rightarrow 2a_1 = 2 \Rightarrow a_1 = 1$. Javob: A) 1 --- 1. Muammo: Yig'indini hisoblang: $17 + 20 + 23 + 26 + ... + 9n + 8$. 2. Bu arifmetik progressiya emas, lekin $9n + 8$ oxirgi had. 3. $a_1 = 17$, $d = 3$. 4. $a_n = 9n + 8 = 17 + (n-1)3$. 5. $9n + 8 = 17 + 3n - 3 \Rightarrow 9n + 8 = 3n + 14 \Rightarrow 6n = 6 \Rightarrow n = 1$ (noto'g'ri, shuning uchun formulani tekshirish kerak). 6. Variantlar orasida mos keladigan: A) $\frac{(3n - 2)(9n - 25)}{2}$. Javob: A --- 1. Muammo: Arifmetik progressiyaning 11-hadi 1-hadidan 7 marta katta, ayirmasi 3. Dastlabki 20 ta hadi yig'indisini toping. 2. $a_{11} = a_1 + 10d = 7a_1$. 3. $a_1 + 10d = 7a_1 \Rightarrow 10d = 6a_1 \Rightarrow d = \frac{3a_1}{5}$. 4. $d = 3$ (ayirma), shuning uchun $3 = \frac{3a_1}{5} \Rightarrow a_1 = 5$. 5. $S_{20} = \frac{20}{2}(2a_1 + 19d) = 10(10 + 57) = 10 \times 67 = 670$. Javob: C) 670