Arifmetik va geometrik progressiya

⭐ — imtihonda eng ko'p ishlatiladigan formulalar. · 🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

Belgilar

• Ketma-ketlik — tartiblangan sonlar: $u_1, u_2, u_3, \dots, u_k, \dots$ ($u_k$ — umumiy had). Qator — ketma-ketlik hadlarining yig'indisi.
• Yig'indi belgisi: $\displaystyle\sum_{k=1}^{5} u_k = u_1 + u_2 + u_3 + u_4 + u_5$

Arifmetik progressiya (A.P.)

Belgilar: $a$ — birinchi had, $d$ — ayirma, $n$ — hadlar soni, $l$ — oxirgi had.

• Keyingi had: $u_{k+1} = u_k + d$
• ⭐ $k$-had: $u_k = a + (k - 1)d$, oxirgi had esa $l = a + (n - 1)d$
• Hadlar soni: $n = \dfrac{l - a}{d} + 1$
• ⭐ Yig'indi: $S_n = \tfrac{1}{2}n\,[2a + (n - 1)d] = \tfrac{1}{2}n\,(a + l)$

Uchta ketma-ket had sharti

$a, b, c$ ketma-ket hadlar bo'lsa:

Yig'indidan hadni topish

$u_n = S_n - S_{n-1}$

🟢 Qoidalar

$\sum_{k=1}^{n}(a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \qquad \sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k \qquad \sum_{k=1}^{n} c = nc$

$\sum_{k=m}^{n} u_k = S_n - S_{m-1}$

🔵 Standart yig'indilar

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2} \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \qquad \sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$

• Birinchi $n$ ta toq son yig'indisi: $1 + 3 + \dots + (2n - 1) = n^2$
• Birinchi $n$ ta juft son yig'indisi: $2 + 4 + \dots + 2n = n(n + 1)$

🟢 $S_n$ — $n$ ga nisbatan kvadrat funksiya

$S_n = \frac{d}{2}\,n^2 + \left(a - \frac{d}{2}\right)n$

Natija: $S_n = pn^2 + qn$ ko'rinishida berilsa, $d = 2p$ va $a = p + q$.

🟢 Hadlar xossasi

• $m + n = p + q$ bo'lsa: $u_m + u_n = u_p + u_q$
• Birinchi va oxirgidan teng uzoqlikdagi hadlar yig'indisi bir xil: $u_1 + u_n = u_2 + u_{n-1} = \dots$

🟢 Qulay belgilashlar

• Uchta ketma-ket had: $a - d,\ a,\ a + d$ (yig'indisi $3a$)
• To'rtta ketma-ket had: $a - 3d,\ a - d,\ a + d,\ a + 3d$ (ayirma $2d$)

Geometrik progressiya (G.P.)

Belgilar: $a$ — birinchi had, $r$ — maxraj (common ratio).

• Keyingi had: $u_{k+1} = r\,u_k$, ya'ni $r = \dfrac{u_{k+1}}{u_k}$
• ⭐ $k$-had: $u_k = a r^{k-1}$, oxirgi had $u_n = a r^{n-1}$
• ⭐ Yig'indi (qatorni $r$ ga ko'paytirib, ayirish usuli bilan topiladi): $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r} \qquad (r \neq 1)$

$r > 1$ bo'lsa birinchisi, $|r| < 1$ bo'lsa ikkinchisi qulayroq.

• Yaqinlashish: $-1 < r < 1$ bo'lsa, qator yaqinlashadi (convergent). $|r| > 1$ bo'lsa uzoqlashadi (divergent), $r = -1$ bo'lsa tebranadi.
• ⭐ Cheksiz G.P. yig'indisi ($-1 < r < 1$): $S_\infty = \frac{a}{1 - r}$

Misol: $8 + 4 + 2 + \dots$ da $a = 8$, $r = \frac{1}{2}$, demak $S_\infty = \dfrac{8}{1 - \frac{1}{2}} = 16$.

• Davriy o'nli kasr — cheksiz G.P.: $0.\dot{2} = 0.2 + 0.02 + 0.002 + \dots = \dfrac{0.2}{1 - 0.1} = \dfrac{2}{9}$

G.P. bo'yicha tez-tez so'raladigan shartlar

• "Yig'indi cheksiz davom etsa ham ma'lum songa yaqinlashadi" ⇔ $|r| < 1$
• $S_\infty$ va $a$ berilsa: $r = 1 - \dfrac{a}{S_\infty}$

🟢 Hadlar xossasi va belgilash

• $m + n = p + q$ bo'lsa: $u_m \cdot u_n = u_p \cdot u_q$
• Uchta ketma-ket had: $\dfrac{a}{r},\ a,\ ar$ (ko'paytmasi $a^3$)
• Birinchi $n$ ta had ko'paytmasi: $a^n\,r^{\frac{n(n-1)}{2}}$

🟢 Davriy o'nli kasrlar

$0.\overline{a} = \frac{a}{9} \qquad 0.\overline{ab} = \frac{ab}{99} \qquad 0.\overline{abc} = \frac{abc}{999}$

Sabab: $0.\overline{45} = 0.45 + 0.0045 + \dots$ — bu $a = 0.45$, $r = 0.01$ bo'lgan cheksiz G.P., $S_\infty = \dfrac{0.45}{0.99} = \dfrac{45}{99}$.

🟢 Murakkab foiz va qadrsizlanish

$A = P\left(1 + \frac{r}{100}\right)^n \qquad A = P\left(1 - \frac{r}{100}\right)^n$

Har yili $r\%$ ga o'sish yoki kamayish — bu maxraji $1 \pm \frac{r}{100}$ bo'lgan G.P.

🟡 Kerakli hadlar sonini topish

$S_n > k$ yoki $u_n < k$ shartidan $n$ ni topish uchun logarifm olinadi:

$r^n > c \;\Rightarrow\; n > \frac{\ln c}{\ln r} \quad (r > 1)$

⚠️ $0 < r < 1$ bo'lsa, $\ln r < 0$ — bo'lganda tengsizlik belgisi teskariga o'zgaradi.

🔵 Arifmetik-geometrik qator

$|r| < 1$ bo'lsa:

$\sum_{k=0}^{\infty} (a + kd)\,r^k = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}$

🟢 Limit

$|r| < 1$ bo'lsa, $n \to \infty$ da $r^n \to 0$. Shuning uchun $S_n \to \dfrac{a}{1 - r}$.