Geometrik progressiya

Bu nima va nega kerak

Tabiat va jamiyatda ba'zi jarayonlar bir maromda emas, balki shiddat bilan karrali ortadi yoki kamayadi. Masalan:

• Laboratoriyada bakteriyalar har 20 daqiqada ikkiga bo'linadi: $1, 2, 4, 8, 16, \ldots$
• Qadimgi shaxmat afsonasida donishmand taxtaning 1-katagiga 1 dona, 2-katagiga 2 dona, 3-katagiga 4 dona bug'doy so'ragan. 64-katakka yetganda bu miqdor $2^{63}$ donaga yetib, butun dunyo hosilidan oshib ketgan.

Har bir keyingi son oldingisini bitta o'zgarmas songa ko'paytirish orqali hosil bo'ladigan bunday ketma-ketlik geometrik progressiya deb ataladi.

Asosiy g'oya

1. Ta'rif va maxraj

Ta'rif. Birinchi hadi $b_1 \neq 0$ bo'lgan, ikkinchi hadidan boshlab har bir hadi o'zidan oldingi hadni bitta noldan farqli o'zgarmas $q$ soniga ko'paytirishdan hosil bo'ladigan ketma-ketlik geometrik progressiya deyiladi: $b_{n+1} = b_n \cdot q \quad (b_1 \neq 0, \; q \neq 0, \; n \in \mathbb{N}).$

$q$ soni geometrik progressiyaning maxraji deb ataladi: $q = \frac{b_{n+1}}{b_n} = \frac{b_2}{b_1} = \frac{b_3}{b_2} = \ldots$

Maxraj $q$ va birinchi had $b_1$ ga qarab:

• $b_1 > 0, q > 1$ bo'lsa — o'suvchi (masalan: $3, 6, 12, 24, \ldots$);
• $b_1 > 0, 0 < q < 1$ bo'lsa — kamayuvchi (masalan: $16, 8, 4, 2, \ldots$);
• $q < 0$ bo'lsa — ishorasi almashinuvchi (masalan: $5, -15, 45, -135, \ldots$);
• $q = 1$ bo'lsa — o'zgarmas (masalan: $4, 4, 4, 4, \ldots$).

2. $n$-had formulasi

Ketma-ketlik hadlarini ta'rif bo'yicha yozsak: $b_1, \quad b_2 = b_1 q, \quad b_3 = b_2 q = b_1 q^2, \quad b_4 = b_1 q^3, \quad \ldots$

Ixtiyoriy $n$-had uchun formula kelib chiqadi: $b_n = b_1 \cdot q^{n-1}.$

Ixtiyoriy ikkita $b_n$ va $b_m$ hadlar orasidagi munosabat: $b_n = b_m \cdot q^{n-m} \implies q^{n-m} = \frac{b_n}{b_m}.$

3. Xarakteristik xossa va indekslar qoidasi

• O'rta geometrik xossasi: Ikkinchi hadidan boshlab har bir hadning kvadrati o'ziga qo'shni hadlar ko'paytmasiga teng: $b_n^2 = b_{n-1}\cdot b_{n+1} \quad (n \ge 2).$ Barcha hadlar musbat bo'lsa: $b_n = \sqrt{b_{n-1}b_{n+1}}$, ya'ni har bir had qo'shnilarining o'rta geometrigiga teng.

• Indekslar yig'indisi tengligi: Agar $k + m = p + q$ bo'lsa, u holda: $b_k \cdot b_m = b_p \cdot b_q.$ Masalan, $b_2 \cdot b_8 = b_4 \cdot b_6 = b_5^2$ (chunki $2 + 8 = 4 + 6 = 10$).

4. Dastlabki $n$ ta had yig'indisi

Geometrik progressiyaning dastlabki $n$ ta hadi yig'indisi $S_n = b_1 + b_2 + \ldots + b_n$ quyidagi formula bilan hisoblanadi:

$S_n = \frac{b_1(q^n - 1)}{q - 1} = \frac{b_1(1 - q^n)}{1 - q} \quad (q \neq 1).$

Oxirgi had $b_n$ ma'lum bo'lganda: $S_n = \frac{b_n q - b_1}{q - 1}.$

Agar $q = 1$ bo'lsa, barcha hadlar teng bo'lib, $S_n = n \cdot b_1$ bo'ladi.

Yechilgan misollar

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

Geometrik progressiyada $b_1 = 48$ va $q = -\dfrac{1}{2}$ berilgan. Uning beshinchi hadi $b_5$ va dastlabki 5 ta hadi yig'indisi $S_5$ ni toping.

Yechish:

1. $n$-had formulasiga $n = 5$ ni qo'yamiz: $b_5 = b_1 \cdot q^4 = 48 \cdot \left(-\frac{1}{2}\right)^4 = 48 \cdot \frac{1}{16} = 3.$
2. Yig'indi formulasidan foydalanamiz: $S_5 = \frac{b_1(1 - q^5)}{1 - q} = \frac{48\left(1 - \left(-\frac{1}{2}\right)^5\right)}{1 - \left(-\frac{1}{2}\right)} = \frac{48\left(1 + \frac{1}{32}\right)}{\frac{3}{2}} = \frac{48 \cdot \frac{33}{32}}{\frac{3}{2}} = \frac{\frac{99}{2}}{\frac{3}{2}} = 33.$

Tekshirish: $48 - 24 + 12 - 6 + 3 = 33$ ✓.

Javob: $b_5 = 3, \quad S_5 = 33$.

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2-misol. Ikki had berilganda progressiyani topish

Hadlari musbat bo'lgan geometrik progressiyada $b_3 = 12$ va $b_6 = 96$ bo'lsa, $b_1$ va $S_6$ ni toping.

Yechish:

1. Maxrajni hadlar nisbatidan topamiz: $\frac{b_6}{b_3} = q^{6-3} = q^3 \implies q^3 = \frac{96}{12} = 8 \implies q = 2.$
2. Birinchi hadni aniqlaymiz: $b_3 = b_1 q^2 \implies 12 = b_1 \cdot 4 \implies b_1 = 3.$
3. Dastlabki 6 ta had yig'indisini hisoblaymiz: $S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(64 - 1) = 3 \cdot 63 = 189.$

Javob: $b_1 = 3, \quad S_6 = 189$.

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3-misol. Tenglamalar sistemasi orqali topish

Geometrik progressiyada $b_1 + b_3 = 15$ va $b_2 + b_4 = 30$ bo'lsa, $b_1$ va $q$ ni toping.

Yechish:

1. Umumiy ko'paytuvchini qavsdan chiqaramiz: $\begin{cases} b_1(1 + q^2) = 15 \\ b_1 q(1 + q^2) = 30 \end{cases}$
2. Ikkinchi tenglikni birinchisiga bo'lamiz: $\frac{b_1 q(1 + q^2)}{b_1(1 + q^2)} = \frac{30}{15} \implies q = 2.$
3. $q = 2$ ni birinchi tenglamaga qo'yamiz: $b_1(1 + 4) = 15 \implies 5b_1 = 15 \implies b_1 = 3.$

Javob: $b_1 = 3, \quad q = 2$.

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4-misol. Ketma-ket uchta had xossasi

$x$ ning qanday musbat qiymatida $x - 1, \; x + 1, \; 2x + 5$ ketma-ketlik geometrik progressiyaning ketma-ket uchta hadi bo'ladi?

Yechish:

1. O'rta geometrik xossasini qo'llaymiz: $(x + 1)^2 = (x - 1)(2x + 5).$
2. Ochib chiqamiz: $x^2 + 2x + 1 = 2x^2 + 3x - 5 \implies x^2 + x - 6 = 0.$
3. Ildizlarni topamiz: $(x + 3)(x - 2) = 0 \implies x_1 = 2, \; x_2 = -3$.
   Musbat qiymat: $x = 2$ (hadlar: $1, 3, 9$).

Javob: $x = 2$.

Mashq

Quyidagi masalalarni mustaqil yeching:

1. Geometrik progressiyada $b_1 = 5, q = 2$ bo'lsa, $b_6$ ni toping.
2. $81, 27, 9, \ldots$ geometrik progressiyaning maxraji $q$ va beshinchi hadi $b_5$ ni toping.
3. Hadlari musbat bo'lgan geometrik progressiyada $b_1 = 2$ va $b_5 = 32$ bo'lsa, maxraj $q$ ni toping.
4. $b_3 = 5$ va $b_7 = 80$ bo'lgan musbat hadli geometrik progressiyada $b_5$ ni toping.
5. $b_1 = 3, q = 2$ bo'lsa, dastlabki 5 ta had yig'indisi $S_5$ ni hisoblang.
6. Hadlari musbat bo'lgan geometrik progressiyada $b_2 = 6, b_4 = 54$. $b_1$ va $S_4$ ni toping.
7. Hadlari musbat geometrik progressiyada $b_1 + b_2 = 12$ va $b_3 + b_4 = 48$ bo'lsa, $q$ va $b_1$ ni toping.
8. $x$ ning qanday musbat qiymatida $2, x, 18$ ketma-ketlik geometrik progressiya hosil qiladi?

Javoblar va yechim yo'llari

1. $b_6 = b_1 q^5 = 5\cdot 32 = 160$.
   Javob: $160$.

2. $q = \dfrac{27}{81} = \dfrac{1}{3}$. $b_5 = 81\cdot \left(\dfrac{1}{3}\right)^4 = 1$.
   Javob: $q = \dfrac{1}{3}, \; b_5 = 1$.

3. $q^4 = \dfrac{b_5}{b_1} = \dfrac{32}{2} = 16 \implies q = 2$.
   Javob: $2$.

4. $b_5^2 = b_3\cdot b_7 = 5\cdot 80 = 400 \implies b_5 = 20$.
   Javob: $20$.

5. $S_5 = \dfrac{3(32 - 1)}{2 - 1} = 3\cdot 31 = 93$.
   Javob: $93$.

6. $q^2 = \dfrac{b_4}{b_2} = \dfrac{54}{6} = 9 \implies q = 3$.
   $b_1 = \dfrac{6}{3} = 2$.
   $S_4 = \dfrac{2(3^4 - 1)}{3 - 1} = 81 - 1 = 80$.
   Javob: $b_1 = 2, \; S_4 = 80$.

7. $q^2 = \dfrac{b_3 + b_4}{b_1 + b_2} = \dfrac{48}{12} = 4 \implies q = 2$.
   $b_1(1 + 2) = 12 \implies b_1 = 4$.
   Javob: $q = 2, \; b_1 = 4$.

8. $x^2 = 2\cdot 18 = 36 \implies x = 6$.
   Javob: $6$.

Keng tarqalgan xatolar

1. $n$-had darajasida $(n - 1)$ o'rniga $n$ ni yozish. ✗ $b_5 = b_1 \cdot q^5$. ✓ Birinchi hadda $q$ qatnashmaydi: $b_5 = b_1 \cdot q^4$.

2. Maxraj manfiy bo'lganda darajani noto'g'ri hisoblash. ✗ $q = -2$ bo'lganda $q^4 = -16$. ✓ Juft darajada manfiy son musbat bo'ladi: $(-2)^4 = 16$.

3. O'rta geometrik xossasida ishorani unutish. ✗ $b_n^2 = 36$ dan faqat $b_n = 6$ deb olish. ✓ Hadlar musbatligi aytilmagan bo'lsa, $b_n = 6$ yoki $b_n = -6$ bo'ladi.

4. Yig'indi maxrajini adashtirish. ✗ $S_n = \dfrac{b_1(q^n - 1)}{q^n - 1}$. ✓ Maxrajda daraja yo'q: $S_n = \dfrac{b_1(q^n - 1)}{q - 1}$.

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Keyingi qadam

Agar geometrik progressiyada $|q| < 1$ bo'lsa, hadlar borgan sari kichiklashib, nolga intiladi. Masalan: $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$. Cheksiz ko'p bunday sonlarni qo'shganda chekli son chiqishi mumkinmi? Keyingi darsda matematikadagi eng ajoyib mavzulardan biri — Cheksiz kamayuvchi geometrik progressiyani o'rganamiz.