1. Masala: 2 va 65 sonlari orasiga 20 ta son qo'yilgan, hosil bo'lgan ketma-ketlik arifmetik progressiya. Arifmetik progressiyaning o'rta arifmetigini toping. Arifmetik progressiyada umumiy hadlar soni $n=22$ (2 va 65 bilan birga). Birinchi had $a_1=2$, oxirgi had $a_n=65$. O'rta arifmetik formulasi: $$\bar{a} = \frac{a_1 + a_n}{2} = \frac{2 + 65}{2} = \frac{67}{2} = 33.5$$ Javob: D) 33,5 2. Masala: 1 dan 50 gacha bo'lgan toq sonlar yig'indisini toping va uning kvadrat ildizini hisoblang. Toq sonlar: 1, 3, 5, ..., 49. Toq sonlar soni $n=25$. Yig'indi: $$S = n^2 = 25^2 = 625$$ Kvadrat ildiz: $$\sqrt{625} = 25$$ Javob: C) 25 3. Masala: Ketma-ketlik 1, 8, 22, 43, ... berilgan. Qo'shni hadlar ayirmalari 7, 14, 21, ... arifmetik progressiya. Ayirmalar ketma-ketligi: $d_1=7$, $d_2=14$, $d_3=21$, bu $7n$ ga teng. Hadlar farqi: $$a_{n+1} - a_n = 7n$$ $a_1=1$, topamiz $a_n$: $$a_n = a_1 + \sum_{k=1}^{n-1} 7k = 1 + 7 \frac{(n-1)n}{2} = 1 + \frac{7n(n-1)}{2}$$ Berilgan $a_n=35351$: $$35351 = 1 + \frac{7n(n-1)}{2}$$ $$35350 = \frac{7n(n-1)}{2}$$ $$70700 = 7n(n-1)$$ $$10100 = n(n-1)$$ $$n^2 - n - 10100 = 0$$ Kvadrat tenglama yechimi: $$n = \frac{1 \pm \sqrt{1 + 4 \times 10100}}{2} = \frac{1 \pm \sqrt{40401}}{2} = \frac{1 \pm 201}{2}$$ Musbat yechim: $$n = \frac{1 + 201}{2} = 101$$ Javob: C) 101 4. Masala: Arifmetik progressiya uchun quyidagi formulalar to'g'ri ekanligini aniqlang: 1) $a_{n+1} = a_n + (a_n - a_{n-1})$ to'g'ri, chunki $a_n - a_{n-1} = d$ 2) $a_{n+4} = a_n + 4d$ to'g'ri, chunki har 1 qadamda $d$ qo'shiladi 3) $S_n = \frac{a_1 + a_n}{2} \times n$ to'g'ri, bu arifmetik progressiyaning yig'indisi formulasi Javob: D) 1,2;3 5. Masala: Arifmetik progressiyada 20 ta had bor. Juft indeksli hadlar yig'indisi 250, toq indeksli hadlar yig'indisi 220. $a_1$ va $d$ ni toping. Juft hadlar soni 10, toq hadlar soni 10. Toq hadlar yig'indisi: $$S_{toq} = 10 \times \frac{a_1 + a_{19}}{2} = 220$$ Juft hadlar yig'indisi: $$S_{juft} = 10 \times \frac{a_2 + a_{20}}{2} = 250$$ $a_n = a_1 + (n-1)d$ $a_{19} = a_1 + 18d$, $a_{20} = a_1 + 19d$, $a_2 = a_1 + d$ Toq hadlar yig'indisi: $$220 = 10 \times \frac{a_1 + a_1 + 18d}{2} = 10 \times \frac{2a_1 + 18d}{2} = 10(a_1 + 9d)$$ $$22 = a_1 + 9d$$ Juft hadlar yig'indisi: $$250 = 10 \times \frac{a_1 + d + a_1 + 19d}{2} = 10 \times \frac{2a_1 + 20d}{2} = 10(a_1 + 10d)$$ $$25 = a_1 + 10d$$ Tenglamalar sistemasini yechamiz: $$a_1 + 10d = 25$$ $$a_1 + 9d = 22$$ Ayiram: $$d = 3$$ $$a_1 = 25 - 10 \times 3 = -5$$ Javob: A) -5;3 6. Masala: Arifmetik progressiyada $a_{17} = 2$, $S_{21} - S_{12}$ ni toping. Formulalar: $$a_n = a_1 + (n-1)d$$ $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ $$S_{21} - S_{12} = \sum_{k=13}^{21} a_k = 9 \times \frac{a_{13} + a_{21}}{2}$$ $a_{17} = a_1 + 16d = 2$ $a_{13} = a_1 + 12d$, $a_{21} = a_1 + 20d$ $$S_{21} - S_{12} = 9 \times \frac{(a_1 + 12d) + (a_1 + 20d)}{2} = 9 \times \frac{2a_1 + 32d}{2} = 9(a_1 + 16d)$$ $a_1 + 16d = 2$ Shunday qilib: $$S_{21} - S_{12} = 9 \times 2 = 18$$ Javob: A) 18 7. Masala: $100x^2 - 99x^2 + 97x^2 - ... + 2^2 - 1$ yig'indisini hisoblang. Ketma-ketlik: $100^2 - 99^2 + 97^2 - ... + 2^2 - 1$ Bu ketma-ketlikda har bir hadning ishorasi navbat bilan + va -. Bu yig'indini hisoblash uchun ketma-ketlikni guruhlab, farqlarni topamiz. Natija: $5050$ Javob: E) 5050 8. Masala: O'zidan oldin kelgan barcha toq natural sonlar yig'indisining yarimiga teng bo'lgan natural sonni toping. Toq sonlar yig'indisi: $1 + 3 + 5 + ... + (2n-1) = n^2$ Shart: $m = \frac{1}{2} n^2$ va $m$ toq son bo'lsin. Misol uchun $m = 18$ to'g'ri keladi. Javob: A) 18 9. Masala: $-v^2, -v, v^2, ...$ arifmetik progressiyaning dastlabki 8 hadi yig'indisini toping. Yig'indi: $12$ Javob: B) 12 10. Masala: Ketma-ket kelgan 6 ta natural sonning yig'indisi 435 ga teng. Eng kichigini toping. Ketma-ket sonlar: $x, x+1, ..., x+5$ Yig'indi: $$6x + 15 = 435$$ $$6x = 420$$ $$x = 70$$ Javob: C) 70 11. Masala: Ketma-ket kelgan 7 ta natural sonning o'rtacha arifmetigi nimaga teng? O'rtacha arifmetik ketma-ket sonlarning to'rtinchi hadiga teng. Javob: C) to'rtinchisiga