Dany jest ciąg geometryczny, taki że a2= √3/2, a3= - 3/2. Wyznacz a1 tego ciągu. Bardzo prosze, pilne :(
q=-3/2:√3/2=-3/√3=-√3
a1*q=a2
a1=√3/2:(-√3)=-1/2
II sposob
a2²=a1*a3
3/4=a1*(-3/2)
a1= -1/2
a2=√3/2
a3=-3/2
q=a3/a2
q=-3/2:√3/2=-3/2 * 2/√3 = -3/√3 = -3*√3/√3*√3=-3√3/3=-√3
a1=a2/q
a1=√3/2:(-√3)=√3/2 * (-1/√3) = -1/2
odp:
a1=-1/2
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q=-3/2:√3/2=-3/√3=-√3
a1*q=a2
a1=√3/2:(-√3)=-1/2
II sposob
a2²=a1*a3
3/4=a1*(-3/2)
a1= -1/2
a2=√3/2
a3=-3/2
q=a3/a2
q=-3/2:√3/2=-3/2 * 2/√3 = -3/√3 = -3*√3/√3*√3=-3√3/3=-√3
a1*q=a2
a1=a2/q
a1=√3/2:(-√3)=√3/2 * (-1/√3) = -1/2
odp:
a1=-1/2