Witajcie :)
Mam do rozwiązania 2 proste zadanka z ciągów, a konkretnie z ciągów geometrycznych
Zadania w załączniku
Proszę o ich dokładne rozwiązania krok po kroku!
Pozdrawiam i daję naj ; ]
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1/
a2=a1*q => q=a2/a1
a) q=64/128= 0,5
wzór ogólny:![a_n=128 * (\frac{1}{2})^{n-1} a_n=128 * (\frac{1}{2})^{n-1}](https://tex.z-dn.net/?f=a_n%3D128+%2A+%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7Bn-1%7D)
a7= a1* q^6
a7= 128 * (1/2)^6 = 2
b)![q= \frac{\frac{1}{4}}{\frac{1}{16}} = 4 q= \frac{\frac{1}{4}}{\frac{1}{16}} = 4](https://tex.z-dn.net/?f=q%3D+%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B16%7D%7D+%3D+4)
c)![q= \frac{-\frac{1}{81}}{\frac{1}{243}} = -3 q= \frac{-\frac{1}{81}}{\frac{1}{243}} = -3](https://tex.z-dn.net/?f=q%3D+%5Cfrac%7B-%5Cfrac%7B1%7D%7B81%7D%7D%7B%5Cfrac%7B1%7D%7B243%7D%7D+%3D+-3)
d)![q= \frac{-\frac{125}{64}}{\frac{625}{256}} = \frac{4}{5} q= \frac{-\frac{125}{64}}{\frac{625}{256}} = \frac{4}{5}](https://tex.z-dn.net/?f=q%3D+%5Cfrac%7B-%5Cfrac%7B125%7D%7B64%7D%7D%7B%5Cfrac%7B625%7D%7B256%7D%7D+%3D+%5Cfrac%7B4%7D%7B5%7D)
e)![q= \frac{-2}{\sqrt2} = -\sqrt2 q= \frac{-2}{\sqrt2} = -\sqrt2](https://tex.z-dn.net/?f=q%3D+%5Cfrac%7B-2%7D%7B%5Csqrt2%7D+%3D+-%5Csqrt2)
f)![q= \frac{\frac{1}{3}}{\frac{2}{9}} = \frac{3}{2} q= \frac{\frac{1}{3}}{\frac{2}{9}} = \frac{3}{2}](https://tex.z-dn.net/?f=q%3D+%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%7D%7B%5Cfrac%7B2%7D%7B9%7D%7D+%3D+%5Cfrac%7B3%7D%7B2%7D)
2/
a) q=-9/-27 = 1/3
-9*1/3=-3
-3*1/3=-1
-27 / 1/3 = -27*3=-81
-81*3= -243
-243,-81,-27,-9,-3,-1
b)
q=5/3
3/ 5/3 = 3*3/5= 9/5
9/5 * 3/5 = 27/25
27/25 * 3/5 = 81/125
81/125 * 3/5 = 243/625
243/625, 81/125, 27/25, 9/5, 3, 5
c) 48=6*q^3
q^3=48/6
q^3=8
q=2
6/2=3
6*2=12
12*2=24
48*2=96
3,6,12,24,48,96
d)q=2/8=1/4
2*1/4=1/2
1/2 * 1/4 = 1/8
1/8 * 1/4 = 1/32
1/32 * 1/4 = 1/128
8,2,1/2, 1/8, 1/32, 1/128, 1/512
e) -81=3*q^3
q^3=-27
q=-3
3/-3= -1
-1/-3 = 1/3
3*-3=-9
-9*-3=27
1/3, -1, 3, -9, 27, -81
f)
54=18*q^2
q^2=3
q=√3 v q=-√3
18*√3=18√3
18/√3= 6√3
6√3 / √3 = 6
6/√3= 2√3
2√3, 6, 6√3, 18, 18√3, 54
lub
-2√3, 6, -6√3, 18, -18√3, 54 (gdy q=-√3)