高二数学,等差数列求解答。
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an=a1+(n-1)d
Sn=a1+a2+...+an
a5/a3 = 5/9
(a1+4d)/(a1+2d) =5/9
9(a1+4d)=5(a1+2d)
4a1=-26d
a1= -(13/2)d (1)
S9/S5
= 9(a1+4d)/[5(a1+2d)]
= 9( -13/2+4)/[5(-13/2+2)]
=9(-5/2)/[ 5(-9/2) ]
=1
Sn=a1+a2+...+an
a5/a3 = 5/9
(a1+4d)/(a1+2d) =5/9
9(a1+4d)=5(a1+2d)
4a1=-26d
a1= -(13/2)d (1)
S9/S5
= 9(a1+4d)/[5(a1+2d)]
= 9( -13/2+4)/[5(-13/2+2)]
=9(-5/2)/[ 5(-9/2) ]
=1
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S9/S5
=[(a1+a9)×9/2]/[(a1+a5)×5/2]
=(2a5×9/2)/(2a3×5/2)
=(9/5)(a5/a3)
=(9/5)(5/9)
=1
等差中项性质:2a5=a1+a9,2a3=a1+a5
等差数列求和公式:Sn=(a1+an)n/2
=[(a1+a9)×9/2]/[(a1+a5)×5/2]
=(2a5×9/2)/(2a3×5/2)
=(9/5)(a5/a3)
=(9/5)(5/9)
=1
等差中项性质:2a5=a1+a9,2a3=a1+a5
等差数列求和公式:Sn=(a1+an)n/2
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等差数列中,a5/a3=5/9,求S9/S5
a5/a3
=(a1+4d)/(a1+2d)=5/9
得到2a1=-13d
S9/S5
=[(a1+a9)*9/2]/[(a1+a5)*5/2]
=[(a1+a9)*9]/[(a1+a5)*5]
=(9/5)*[(2a1+8d)/(2a1+4d)]
=(9/5)*(-5d/-9d)
=1,即为所求
a5/a3
=(a1+4d)/(a1+2d)=5/9
得到2a1=-13d
S9/S5
=[(a1+a9)*9/2]/[(a1+a5)*5/2]
=[(a1+a9)*9]/[(a1+a5)*5]
=(9/5)*[(2a1+8d)/(2a1+4d)]
=(9/5)*(-5d/-9d)
=1,即为所求
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