Odpowiedź:
Zad 1
a1 = -5 + 3 = -2
a2 = -10 + 3 = -7
a3 = -15 + 3 = -12
a4 = -20 + 3 = -17
Jest arytmetyczny ponieważ ma stałą różnicę równą -5, ciąg ten jest malejący.
Zad 2
a1 = 5
a8 = 26
a8 = a1 + 7r
26 = 5 + 7r
21 = 7r
r = 3
Zad 3
a1 = 12
a2 = 8 r = -4
a3 = 4
a4 = 0
a5 = -4
a6 = -8
a7 = -12
S(50) = (a1 + a50) /2 • 50
a50 = a1 + 49r = 12 + 49 • (-4) = 12 - 196 = - 184
S(50) = (12 - 184) /2 • 50
S(50) = -86 • 50 = - 4300
S(50) = -4300
Zad 4
1,8,15,22,29,36,43 r = 7
Zad 5
a3 + a7 = 2
a8 - a5 = -9
a1 + 2r + a1 + 6r = 2
2a1 + 8r = 2
a1 + 7r - (a1 + 4r) = -9
a1 + 7r - a1 - 4r = -9
3r = -9
r = -3
2a1 - 24 = 2
2a1 = 26
a1 = 13
Zad 6
2(x²+3) = 2 + x + 10
2x² + 6 = 2 + x + 10
2x² - x - 6 = 0
∆ = (-1)² - 4 • 2 • -6 = 1 + 48 = 49
√∆ = 7
x1 = (1 - 7) /4 = -6/4 = -3/2
x2 = (1+7) /4 = 8/4 = 2
2,7,12 r = 5
2, 5,25 , 8,5 r = 3,25
Zad 7
an = 99
an = a1 + (n-1) • r
99 = 12 + (n-1) • 3
99 = 12 + 3n - 3
99 - 9 = 3n
90 = 3n
n = 30
S(30) = (12 + 99) /2 • 30
S(30) = 1845
Zad 8
Sn = [2a1 + (n - 1) • r]/2 • n
945 = [2 • 10 + 5n - 5]/2 • n
945 = [20 + 5n - 5] /2 • n // •2
1890 = 20n + 5n² - 5n
5n² + 15n - 1890 = 0 // :5
n² + 3n - 378 = 0
∆ = 9 - 4 • 1 • (-378) = 9 + 1512 = 1521
√∆ = 39
x1 = (-3 - 39) /2 = - 42 /2 = -21 ta nie będzie bo liczba wyrazów musi być dodatnią
x2 = (-3 + 39) /2 = 36/2 = 18
18 początkowych wyrazów
Już jest wszystko dobrze.
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Odpowiedź:
Zad 1
a1 = -5 + 3 = -2
a2 = -10 + 3 = -7
a3 = -15 + 3 = -12
a4 = -20 + 3 = -17
Jest arytmetyczny ponieważ ma stałą różnicę równą -5, ciąg ten jest malejący.
Zad 2
a1 = 5
a8 = 26
a8 = a1 + 7r
26 = 5 + 7r
21 = 7r
r = 3
Zad 3
a1 = 12
a2 = 8 r = -4
a3 = 4
a4 = 0
a5 = -4
a6 = -8
a7 = -12
S(50) = (a1 + a50) /2 • 50
a50 = a1 + 49r = 12 + 49 • (-4) = 12 - 196 = - 184
S(50) = (12 - 184) /2 • 50
S(50) = -86 • 50 = - 4300
S(50) = -4300
Zad 4
1,8,15,22,29,36,43 r = 7
Zad 5
a3 + a7 = 2
a8 - a5 = -9
a1 + 2r + a1 + 6r = 2
2a1 + 8r = 2
a1 + 7r - (a1 + 4r) = -9
a1 + 7r - a1 - 4r = -9
3r = -9
r = -3
2a1 - 24 = 2
2a1 = 26
a1 = 13
Zad 6
2(x²+3) = 2 + x + 10
2x² + 6 = 2 + x + 10
2x² - x - 6 = 0
∆ = (-1)² - 4 • 2 • -6 = 1 + 48 = 49
√∆ = 7
x1 = (1 - 7) /4 = -6/4 = -3/2
x2 = (1+7) /4 = 8/4 = 2
2,7,12 r = 5
2, 5,25 , 8,5 r = 3,25
Zad 7
a1 = 12
an = 99
an = a1 + (n-1) • r
99 = 12 + (n-1) • 3
99 = 12 + 3n - 3
99 - 9 = 3n
90 = 3n
n = 30
S(30) = (12 + 99) /2 • 30
S(30) = 1845
Zad 8
Sn = [2a1 + (n - 1) • r]/2 • n
945 = [2 • 10 + 5n - 5]/2 • n
945 = [20 + 5n - 5] /2 • n // •2
1890 = 20n + 5n² - 5n
5n² + 15n - 1890 = 0 // :5
n² + 3n - 378 = 0
∆ = 9 - 4 • 1 • (-378) = 9 + 1512 = 1521
√∆ = 39
x1 = (-3 - 39) /2 = - 42 /2 = -21 ta nie będzie bo liczba wyrazów musi być dodatnią
x2 = (-3 + 39) /2 = 36/2 = 18
18 początkowych wyrazów
Już jest wszystko dobrze.