Geometrik progressiya

Ta'rif

Geometrik progressiya (GP) shunday ketma-ketlik, unda ikkinchi hadidan boshlab har bir had oldingisini bir xil q soniga ko'paytirishdan hosil bo'ladi (b₁ ≠ 0, q ≠ 0). q maxraj: q = bₙ₊₁ / bₙ.

• Misol: 3, 6, 12, 24, ... bunda b₁ = 3, q = 2.
• |q| > 1 bo'lsa hadlar modul bo'yicha o'sadi, |q| < 1 bo'lsa kamayadi, q < 0 bo'lsa ishoralar almashadi.

Asosiy formulalar

Nima topiladi	Formula
n-had	bₙ = b₁ · qⁿ⁻¹
Dastlabki n had yig'indisi (q ≠ 1)	Sₙ = b₁ · (qⁿ − 1) / (q − 1)
q = 1 bo'lsa	Sₙ = n · b₁
Xarakteristik xossa	bₙ² = bₙ₋₁ · bₙ₊₁
Cheksiz kamayuvchi GP (−1 < q < 1)	S = b₁ / (1 − q)

• Ketma-ket uch musbat son a, b, c GP tashkil qilsa, b² = ac.
• bₙ = bₘ · qⁿ⁻ᵐ. Ikki had berilsa, qⁿ⁻ᵐ = bₙ / bₘ.
• Indekslar yig'indisi teng bo'lsa, hadlar ko'paytmasi teng: k + m = p + r bo'lsa, bₖ · bₘ = bₚ · bᵣ.

Yechish tartibi

1. b₁ va q ni toping (kerak bo'lsa bₙ / bₘ dan)
2. Kerakli formulani tanlang: bₙ, Sₙ yoki cheksiz yig'indi
3. Qiymatlarni qo'yib hisoblang
4. Natijani shartga solib tekshiring

Misollar

• b₁ = 2, q = 3 bo'lsa, b₄ = 2 · 27 = 54 va S₄ = 2 · (81 − 1) / 2 = 80.
• b₂ = 6, b₅ = 162: q³ = 162 / 6 = 27, demak q = 3 va b₁ = 2.
• 6, 2, 2/3, ... da q = 1/3, S = 6 / (1 − 1/3) = 9.

Xatolar

• Cheksiz yig'indi faqat −1 < q < 1 da.
• Ko'rsatkich n − 1 bo'ladi, n emas: b₅ = b₁ · q⁴.
• Maxrajni topishda q = keyingi had / oldingi had.