Question

1. W ciągu arytmetycznym suma wyrazów drugiego i trzeciego wynosi 8, zaś suma wyrazów drugiego i piątego wynosi 14. Oblicz wyrazy a4 i a10.

a1=
r=
a4=
a10=

2. W ciągu geometrycznym a3 = 50 i a4 = 250. Oblicz a2 i S4.

a1=
q=
a2=
S4=

Answer 1

1.
a2+a3 = 8 -> a1+r+a1+2r=8 ->> 2a1 + 3r = 8
a2+a5=14 -> a1+r+a1+4r=14 ->> 2a1+5r =14

2a1+5r =14
2a1+3r=8
2a1=14-5r
2a1=8-3r
14-5r=8-3r
-2r=-6r
r=3

2a1+3r=8
2a1=8-3*3
2a1=-1
a1 = -1/2

a1= -1/2
r= 3
a4= a1+3r = -1/2+3*3=-1/2+9=8 i 1/2
a10= a1+9r=-1/2+9*3=-1/2+27=26 i 1/2



2.
a3 = 50
a4 = 250
a3 = a1*q² ->> a1 = a3 / q² = 50 / q²
a4 = a1*q³ ->> a1= a4/q³ = 250 / q³

a1 = 50 / q²
a1 = 250 / q³

50 / q² = 250 / q³ /*q³
50q=250
q=5

a1 = 50 / q² = 50 / 5² = 50/25 = 2

a1= 2
q= 5
a2= a1 * q = 2*5 = 10
S4= a1 (1-q do czwartej) / 1-q = 2*(1-625) / 1-5 = 312