Arifmetik progressiya

Bu nima va nega kerak

Bir maromda o'sib yoki kamayib boruvchi sonlar ketma-ketligiga hayotda tez-tez duch kelamiz. Masalan, teatr zalining birinchi qatorida 20 ta o'rindiq bo'lib, har keyingi qatorda oldingisidan 2 tadan ko'p o'rindiq joylashtirilgan bo'lsa: $20, 22, 24, 26, \ldots$ ketma-ketligi hosil bo'ladi.

Har bir had oldingisiga bitta o'zgarmas sonni qo'shish orqali hosil bo'ladigan bunday ketma-ketlik arifmetik progressiya deyiladi.

Rivoyatga ko'ra, mashhur matematik Karl Gauss bolaligida $1$ dan $100$ gacha sonlar yig'indisini juftlash usulida soniyalarda topgan: $1 + 100 = 101, \quad 2 + 99 = 101, \quad \ldots, \quad 50 + 51 = 101.$ Bunday 50 ta juftlik yig'indisi $50 \times 101 = 5050$ bo'ladi. Arifmetik progressiya formulalari bu g'oyani har qanday qadamli qatorlar uchun umumlashtiradi.

Asosiy g'oya

1. Ta'rif va ayirma

Ta'rif. Ikkinchi hadidan boshlab har bir hadi o'zidan oldingi hadga bitta o'zgarmas $d$ sonini qo'shishdan hosil bo'ladigan ketma-ketlik arifmetik progressiya deyiladi: $a_{n+1} = a_n + d \quad (n \in \mathbb{N}).$

Bu yerdagi $d$ soni arifmetik progressiyaning ayirmasi deyiladi: $d = a_{n+1} - a_n = a_2 - a_1 = a_3 - a_2.$

• Agar $d > 0$ bo'lsa, progressiya o'suvchi (masalan: $2, 5, 8, 11, \ldots$);
• Agar $d < 0$ bo'lsa, progressiya kamayuvchi (masalan: $20, 16, 12, 8, \ldots$);
• Agar $d = 0$ bo'lsa, progressiya o'zgarmas bo'ladi (masalan: $7, 7, 7, 7, \ldots$).

2. $n$-had formulasi

Ketma-ketlik hadlarini yozsak: $a_1, \quad a_2 = a_1 + d, \quad a_3 = a_1 + 2d, \quad \ldots$

Ixtiyoriy $n$-had uchun formula kelib chiqadi: $a_n = a_1 + (n - 1)d.$

Ixtiyoriy ikkita $a_n$ va $a_m$ hadlar orasidagi bog'lanish: $a_n = a_m + (n - m)d \implies d = \frac{a_n - a_m}{n - m}.$

3. Xarakteristik xossa va indekslar qoidasi

• O'rta arifmetik xossasi: Ikkinchi hadidan boshlab har bir had qo'shnilarining o'rta arifmetigiga teng: $a_n = \frac{a_{n-1} + a_{n+1}}{2} \quad (n \ge 2).$

• Indekslar yig'indisi tengligi: Agar $k + m = p + q$ bo'lsa, u holda: $a_k + a_m = a_p + a_q.$ Masalan, $a_2 + a_8 = a_4 + a_6 = 2a_5$ (chunki $2 + 8 = 4 + 6 = 10$).

4. Dastlabki $n$ ta had yig'indisi formulasi

Dastlabki $n$ ta had yig'indisi $S_n = a_1 + a_2 + \ldots + a_n$ quyidagicha hisoblanadi:

1. Birinchi va $n$-had ma'lum bo'lganda: $S_n = \frac{a_1 + a_n}{2}\cdot n.$

2. Faqat $a_1$ va $d$ ma'lum bo'lganda: $S_n = \frac{2a_1 + (n - 1)d}{2}\cdot n.$

Shuningdek, yig'indidan hadni topish qoidasi: $a_n = S_n - S_{n-1} \quad (n \ge 2), \qquad a_1 = S_1.$

Yechilgan misollar

1-misol. $n$-had va yig'indini topish

Arifmetik progressiyada $a_1 = 7$ va $d = -3$ berilgan. Uning $a_{15}$ hadi va dastlabki 15 ta hadi yig'indisi $S_{15}$ ni toping.

Yechish:

1. $n$-had formulasiga $n = 15$ ni qo'yamiz: $a_{15} = a_1 + 14d = 7 + 14\cdot (-3) = 7 - 42 = -35.$
2. Yig'indi formulasidan foydalanamiz: $S_{15} = \frac{a_1 + a_{15}}{2}\cdot 15 = \frac{7 + (-35)}{2}\cdot 15 = \frac{-28}{2}\cdot 15 = -14\cdot 15 = -210.$

Javob: $a_{15} = -35, \quad S_{15} = -210$.

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2-misol. Ikki had berilganda progressiyani topish

Arifmetik progressiyada $a_5 = 17$ va $a_{11} = 35$ bo'lsa, $a_1$, $d$ va $S_{10}$ ni toping.

Yechish:

1. Ayirmani hadlar ayirmasi orqali topamiz: $a_{11} - a_5 = (11 - 5)d \implies 6d = 35 - 17 = 18 \implies d = 3.$
2. Birinchi hadni aniqlaymiz: $a_5 = a_1 + 4d \implies 17 = a_1 + 4\cdot 3 \implies a_1 = 17 - 12 = 5.$
3. Dastlabki 10 ta had yig'indisini hisoblaymiz: $S_{10} = \frac{2a_1 + 9d}{2}\cdot 10 = (2\cdot 5 + 9\cdot 3)\cdot 5 = (10 + 27)\cdot 5 = 37\cdot 5 = 185.$

Javob: $a_1 = 5, \quad d = 3, \quad S_{10} = 185$.

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3-misol. Hadning tartib raqamini topish

$9, 13, 17, \ldots$ arifmetik progressiyaning nechanchi hadi $89$ ga teng?

Yechish:

1. Ketma-ketlikdan: $a_1 = 9, \; d = 13 - 9 = 4, \; a_n = 89$.
2. Formulaga qo'yamiz: $a_n = a_1 + (n - 1)d \implies 89 = 9 + (n - 1)\cdot 4.$ $80 = 4(n - 1) \implies n - 1 = 20 \implies n = 21.$

Javob: 21-hadi.

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4-misol. Yig'indiga ko'ra hadlar sonini aniqlash

Arifmetik progressiyada $a_1 = 3$ va $d = 2$ bo'lsa, nechta hadning yig'indisi $80$ ga teng bo'ladi?

Yechish:

1. Yig'indi formulasiga qo'yamiz: $S_n = \frac{2\cdot 3 + (n - 1)\cdot 2}{2}\cdot n = 80.$ $\frac{6 + 2n - 2}{2}\cdot n = 80 \implies \frac{2n + 4}{2}\cdot n = 80 \implies (n + 2)n = 80.$
2. Kvadrat tenglamani yechamiz: $n^2 + 2n - 80 = 0 \implies (n + 10)(n - 8) = 0.$ Hadlar soni natural son bo'lgani uchun: $n = 8$.

Javob: 8 ta had.

Mashq

Quyidagi masalalarni mustaqil yeching:

1. Arifmetik progressiyada $a_1 = 4, d = 5$ bo'lsa, $a_9$ ni toping.
2. $35, 31, 27, \ldots$ progressiyaning ayirmasi $d$ va 10-hadi $a_{10}$ ni toping.
3. Arifmetik progressiyada $a_1 = -2$ va $a_7 = 16$ bo'lsa, ayirma $d$ ni toping.
4. $a_3 = 8$ va $a_9 = 26$ bo'lgan arifmetik progressiyaning oltinchi hadi $a_6$ ni toping.
5. $2, 6, 10, \ldots$ arifmetik progressiyaning dastlabki 12 ta hadi yig'indisi $S_{12}$ ni hisoblang.
6. Arifmetik progressiyada $a_1 = 10, d = -2$ bo'lsa, dastlabki nechta hadning yig'indisi $24$ ga teng bo'ladi?
7. Dastlabki 20 ta juft natural sonning yig'indisini hisoblang ($2 + 4 + 6 + \ldots + 40$).
8. Dastlabki $n$ ta hadi yig'indisi $S_n = 2n^2 - n$ formula bilan berilgan ketma-ketlikning uchinchi hadi $a_3$ ni toping.

Javoblar va yechim yo'llari

1. $a_9 = a_1 + 8d = 4 + 8\cdot 5 = 44$.
   Javob: $44$.

2. $d = 31 - 35 = -4$. $a_{10} = 35 + 9\cdot (-4) = -1$.
   Javob: $d = -4, \; a_{10} = -1$.

3. $a_7 - a_1 = 6d \implies 18 = 6d \implies d = 3$.
   Javob: $3$.

4. Indekslar xossasidan: $a_6 = \dfrac{a_3 + a_9}{2} = \dfrac{8 + 26}{2} = 17$.
   Javob: $17$.

5. $a_1 = 2, d = 4$. $S_{12} = \dfrac{2\cdot 2 + 11\cdot 4}{2}\cdot 12 = 48\cdot 6 = 288$.
   Javob: $288$.

6. $S_n = \dfrac{20 + (n - 1)(-2)}{2}\cdot n = (11 - n)n = 24 \implies n^2 - 11n + 24 = 0$.
   $(n - 3)(n - 8) = 0 \implies n = 3$ yoki $n = 8$.
   Javob: $3$ yoki $8$ ta had.

7. $a_1 = 2, d = 2, n = 20, a_{20} = 40$.
   $S_{20} = \dfrac{2 + 40}{2}\cdot 20 = 42\cdot 10 = 420$.
   Javob: $420$.

8. $a_3 = S_3 - S_2 = (2\cdot 9 - 3) - (2\cdot 4 - 2) = 15 - 6 = 9$.
   Javob: $9$.

Keng tarqalgan xatolar

1. $n$-had formulasida $(n - 1)$ o'rniga $n$ ni yozish. ✗ $a_{10} = a_1 + 10d$. ✓ $10$-hadda $d$ soni $9$ marta qo'shiladi: $a_{10} = a_1 + 9d$.

2. Ayirma manfiy bo'lganda qavslarni unutish. ✗ $a_1 = 10, d = -3$ bo'lganda $a_5 = 10 + 4\cdot -3$. ✓ Ko'paytirishda manfiy son qavsga olinadi: $a_5 = 10 + 4\cdot (-3) = -2$.

3. Indekslar bilan hadlar qiymatini adashtirish. ✗ $a_3 + a_5 = 20$ dan $a_8 = 20$ deb xulosa qilish. ✓ Indekslar xossasidan $a_3 + a_5 = 2a_4 = 20$, demak $a_4 = 10$.

4. Yig'indi formulasida 2 ga bo'lishni unutish. ✗ $S_n = (a_1 + a_n)\cdot n$. ✓ O'rta arifmetik olingani uchun $2$ ga bo'linadi: $S_n = \dfrac{a_1 + a_n}{2}\cdot n$.

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Keyingi qadam

Arifmetik progressiyada har bir had oldingisiga ma'lum sonni qo'shish orqali hosil bo'ldi. Agar sonlar har safar bitta songa ko'paytirib borilsa-chi? Masalan, hujayralar bo'linishi ($1, 2, 4, 8, \ldots$) qanday qonuniyatga bo'ysunadi? Buni keyingi darsimiz — Geometrik progressiya mavzusida o'rganamiz.