在等差数列{an}中,S7=42,Sn=510,第n-3项等于45,求n
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在等差数列{an}中,S7=42,Sn=510,第n-3项等于45,求n
an=a1+(n-1)d
S7=42
(a1+3d)7=42
a1+3d=6 (1)
Sn =510
(2a1+(n-1)d)n/2 = 510
(2a1+(n-1)d)n = 1020 (2)
a(n-3) =45
a1+(n-4)d =45 (3)
(3)-(1)
(n-7)d = 39 (4)
sub (1) into (2)
(12-6d+(n-1)d)n = 1020
(12+(n-7)d)n = 1020
(12+39)n = 1020 ( from (3))
51n =1020
n =20
再问: 我就是这样列的啊。。。可是,好吧,把7写成5了。。。还是谢谢啦!
S7=42
(a1+3d)7=42
a1+3d=6 (1)
Sn =510
(2a1+(n-1)d)n/2 = 510
(2a1+(n-1)d)n = 1020 (2)
a(n-3) =45
a1+(n-4)d =45 (3)
(3)-(1)
(n-7)d = 39 (4)
sub (1) into (2)
(12-6d+(n-1)d)n = 1020
(12+(n-7)d)n = 1020
(12+39)n = 1020 ( from (3))
51n =1020
n =20
再问: 我就是这样列的啊。。。可是,好吧,把7写成5了。。。还是谢谢啦!
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