Odpowiedź:
z.1
a3 = 9 a8 = 24
więc a8 = a3 + 5 r
24 = 9 +5 r ⇒ 5 r =24 - 9 = 15 / : 5
r = 3
a3 = a1 +2 r
9 = a1 + 2*3
a1 = 9 - 6 = 3
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a5 = a1 +4 r = 3 + 4*3 = 15
S5 = [tex]\frac{3 + 15}{2}[/tex] *5 = 9*5 = 45
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z.2
( x - 5, 3 x + 4, 4 x + 1) - c. arytmetyczny
więc
2*(3 x + 4) = ( x -5) + ( 4 x + 1)
6 x + 8 = 5 x - 4
x = - 12
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więc
a1 = - 12 - 5 = - 17
a2 = 3*(-12) + 4 = - 32
a3 = 4*(-12) + 1 = - 47
Różnica r = - 32 - ( - 17) = - 32 + 17 = - 15
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z.3
a3 = 9
S3 = a1 + a2 +a 3 = 15
S3 = [tex]\frac{a1 +a3}{2}[/tex]* 3 = 15 / : 3
[tex]\frac{a1 + a3}{2}[/tex] = 5 ⇒ a2 = 5
r = a3 -a2 = 9 - 5 = 4
a1 = a2 - r = 5 - 4 = 1
więc
an = a1 + ( n -1)* r = 1 + ( n -1)*4 = 1 + 4 n - 4 = 4 n - 3
an = 4 n - 3
dlatego
Sn = [tex]\frac{a1 + an}{2}[/tex] * n = 780 / * 2
( 1 + 4 n - 3)*n = 780*2
( 4 n - 2)*n = 1560
4 n² - 2 n - 1 560 = 0 / : 2
2 n² - n - 780 = 0
Δ = 1 - 4*2*( -780) = 1 + 6 240 = 6 241
√Δ = 79
n = [tex]\frac{1 + 79}{4}[/tex] = 20
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Trzeba dodać 20 początkowych wyrazów tego ciągu.
Szczegółowe wyjaśnienie:
