Ciąg (a,b,c) jest arytmetyczny i a+b+c=33 Ciąg (a,b+3,c+13) jest geometryczny . oblicz a,b,c
b-a = c-b
2b = c+a
-c-a+2b = 0
a+b+c=33
sumujemy:
3b=33
b=11
(b+3)/a = (c+13)/(b+3)
a*(c+13)= 14*14
dostajemy:
a+c=22
a(c+13)= 196
a=22-c
(22-c)(c+13)=196
22c + 286 -c^2 - 13c = 196
-c^2 + 9c + 90 = 0
Δ = 81 - (-360) = 441
√Δ = 21
c₁=-9-21/-2 = 15
c₂=-9+21/-2 = -6
Wtedy:
a₁=7
a₂=28
Rozwiązania:
a=7
c=15
lub
a=28
c=-6
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b-a = c-b
2b = c+a
-c-a+2b = 0
a+b+c=33
sumujemy:
3b=33
b=11
(b+3)/a = (c+13)/(b+3)
a*(c+13)= 14*14
dostajemy:
a+c=22
a(c+13)= 196
a=22-c
(22-c)(c+13)=196
22c + 286 -c^2 - 13c = 196
-c^2 + 9c + 90 = 0
Δ = 81 - (-360) = 441
√Δ = 21
c₁=-9-21/-2 = 15
c₂=-9+21/-2 = -6
Wtedy:
a₁=7
a₂=28
Rozwiązania:
a=7
b=11
c=15
lub
a=28
b=11
c=-6