Geometrik progressiya
Javoblar ota-onalar va o‘qituvchilar uchun.
1
2, 10, 50, ... progressiyaning maxrajini va b₆ ni toping.
q = 5, b₆ = 6250
2
b₂ = 6, b₅ = 162 bo‘lsa, q va b₁ ni toping.
q = 3, b₁ = 2
3
640 soni 5, 10, 20, ... progressiyaning hadimi? Nomerini toping.
Ha: 5 · 2^(n−1) = 640, 2^(n−1) = 128 = 2⁷, n = 8.
4
b_n = 3 · 4ⁿ ketma-ketlik geometrik progressiya ekanini isbotlang.
b_(n+1) : b_n = (3 · 4^(n+1)) : (3 · 4ⁿ) = 4, n ga bog‘liq emas.
5
4 va 25 sonlarining musbat o‘rta geometrigini toping.
10