1. rozwiąż równanie:
1+7+13+19+...+x=280
2. Oblicz sumę 8 wyrazów ciagu geometrycznego, w którym
1)
x=280-1-7-13-19
x=240
1]
a₁=1
a₂=7
r=7-1=6
x= n-ty wyraz ciagu
x=a₁+(n-1)r
x=1(n-1)6
x=6n-6
280=[ a₁+an]/2 * n
560=[ 1+6n-6] n
560=[6n-5] n
6n²-5n-560=0
Δ=b²-4ac=25+13440=13465
280=[2a₁+(n-1)r]/2 * n
280=[2+(n-1)6]/2 * n
560=[2+6n-6]n
560=6n²-4n
6n²-4n-560=0
Δ=b²-4ac=16+13440=13456
√Δ=116
n₁=[4-116]/12= sprzeczne bo n ∈ N
n=[4+116]/12=10
x=a₁₀=a₁+9r=1+9*6=55
2]
a₆=a₂q⁴
243=3q⁴
q⁴=243:3
q⁴=81
q=⁴√81=3 lub q=-3
dla q=3 a₂=a₁q 3=3a₁ a₁=1
S₈=1(1-q⁸) / (1-q)= (1-3⁸)/(1-3)=(1-6561)-2=3280
dla q=-3 3=-3a₁ a₁=-1
S₈= - 1 (1-(-3)⁸) / (1-(-3)]=-1 * (-6560) /4=1640
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1)
x=280-1-7-13-19
x=240
1]
a₁=1
a₂=7
r=7-1=6
x= n-ty wyraz ciagu
x=a₁+(n-1)r
x=1(n-1)6
x=6n-6
280=[ a₁+an]/2 * n
560=[ 1+6n-6] n
560=[6n-5] n
6n²-5n-560=0
Δ=b²-4ac=25+13440=13465
280=[2a₁+(n-1)r]/2 * n
280=[2+(n-1)6]/2 * n
560=[2+6n-6]n
560=6n²-4n
6n²-4n-560=0
Δ=b²-4ac=16+13440=13456
√Δ=116
n₁=[4-116]/12= sprzeczne bo n ∈ N
n=[4+116]/12=10
x=a₁₀=a₁+9r=1+9*6=55
2]
a₆=a₂q⁴
243=3q⁴
q⁴=243:3
q⁴=81
q=⁴√81=3 lub q=-3
dla q=3 a₂=a₁q 3=3a₁ a₁=1
S₈=1(1-q⁸) / (1-q)= (1-3⁸)/(1-3)=(1-6561)-2=3280
dla q=-3 3=-3a₁ a₁=-1
S₈= - 1 (1-(-3)⁸) / (1-(-3)]=-1 * (-6560) /4=1640