z.1 a3 = 4 oraz a8 = 256 Mamy a3 = a1*q^2 oraz a8 = a1*q^7 a8 : a3 = [a1*q^7] : [ a1*q^2 ] = q^5 ale a8 : a3 = 256 : 4 = 64 q^5 = 64 q = p 5st (64) a1*q^2 = 4 ---> a1 = 4 : q^2 = 4 : [p 5st (64)]^2 = 4 : p5st(4096) =============== z.2 a2 = 4 a6 = 20 a2 = a1 + r a6 = a1 + 5r a6 - a2 = [a1 +5r] - [a1 + r] = 4r ale a6 - a2 = 20 - 4 = 16 czyli 4r = 16 / : 4 r = 4 ==== a2 = a1 + r ---> a1 = a2 - r = 4 - 4 = 0 Odp. a1 = 0 oraz r = 4 ===========================
1) [latex]\a_n=q^{n-1}a_1[/latex] 4=q²a₁ 256=q⁷a₁ a₁=4/q² 256=q⁷a₁ 256=q⁷(4/q²) 256=4q⁵ q⁵=64 q=⁵√64 q=64^(1/5) q=(2⁶)^(1/5) q=2^(6/5) a₁=4/(2^(6/5))² a₁=4/(2^(12/5)) a₁=2²/(2^(12/5)) a₁=2^(2-12/5) a₁=2^(-2/5) an=(2^(6/5))^(n-1)*2^(-2/5) an=2^((6n-6)/5)*2^(-2/5) an=2^((6n-8)/5) 2) [latex]\a_n=a_1+(n-1)r[/latex] 4=a₁+(2-1)r 20=a₁+(6-1)r 4=a₁+r |*-1 20=a₁+5r -4=-a₁-r 20=a₁+5r ---------- 16=4r r=4 4=a₁+4 a₁=0 an=(n-1)*4 an=4n-4
