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Geometrik progressiya
maktab darsligi: algebra 9-sinf, geometrik progressiya
Maqsad: Geometrik progressiyaning n-hadi va yig‘indisi formulalarini ishlatish.
Qoida va g‘oya
Geometrik progressiyada har bir had oldingisini bir xil q songa (maxraj) ko‘paytirib hosil qilinadi. n-had: bn = b1 · q^(n-1). Xossa: b(n)^2 = b(n-1) · b(n+1); musbat hadlarda o‘rtadagi had qo‘shnilarining o‘rta geometrigi. Yig‘indi: Sn = b1 · (q^n - 1)/(q - 1), q ≠ 1. Agar q = 1 bo‘lsa, Sn = n · b1. q < 0 bo‘lsa, hadlar ishorasi navbatma-navbat almashadi. Ikki had berilsa, q^(m-n) = bm/bn. Juft darajali ildiz olishda ikkita q chiqishi mumkin: shartga qarang (hadlar musbat yoki yo‘q).
Namunaviy masalalar
b1 = 2, q = 3 bo‘lsa, b5 ni toping.
Yechim: b5 = 2 · 3^4 = 162.
Javob: 162
Yechim: b5 = 2 · 3^4 = 162.
Javob: 162
1, 2, 4, ... progressiyaning dastlabki 6 ta hadi yig‘indisini toping.
Yechim: S6 = 1 · (2^6 - 1)/(2 - 1) = 63.
Javob: 63
Yechim: S6 = 1 · (2^6 - 1)/(2 - 1) = 63.
Javob: 63
b2 = 6 va b5 = 162 bo‘lsa, b4 ni toping.
Yechim: q^3 = 162/6 = 27, q = 3. b4 = b2 · q^2 = 6 · 9 = 54.
Javob: 54
Yechim: q^3 = 162/6 = 27, q = 3. b4 = b2 · q^2 = 6 · 9 = 54.
Javob: 54
Tipik xatolar
- bn formulasida q^n ni yozish: ko‘rsatkich n - 1 bo‘lishi kerak.
- q^2 = 4 dan q = 2 deb, q = -2 ni shartga qarab tashlashni unutish.
- Geometrik progressiya yig‘indisini arifmetik formula bilan hisoblash.
Mashq
1
b1 = 64, q = 1/2 bo‘lsa, b7 ni toping.
1
2
b1 = 3, q = 2 bo‘lsa, dastlabki 5 ta had yig‘indisini toping.
93
3
4, x, 16 musbat hadli geometrik progressiya. x ni toping.
8
4
b3 = 20, b5 = 80, q > 0 bo‘lsa, b1 ni toping.
5