www.1862.net > 已知A3=3/2,S3=9/2.求A1与q.

已知A3=3/2,S3=9/2.求A1与q.

a3/q² + a3/q + a3 = S3 3/2(1/q² + 1/q + 1) = 9/2 1/q² + 1/q + 1 = 3 1/q² + 1/q - 2 = 0 (1/q - 1)(1/q + 2) = 0 q = 1 或 q = -1/2 当 q = 1 时 ,a1 = 3/2 当 q = -1/2 时 ,a1 = 6

a3/q² + a3/q + a3 = S3 3/2(1/q² + 1/q + 1) = 9/2 1/q² + 1/q + 1 = 3 1/q² + 1/q - 2 = 0 (1/q - 1)(1/q + 2) = 0 解出: q = 1 或 q = -1/2 当 q = 1 时 ,a1 = 3/2 当 q = -1/2 时 ,a1 = 6 希望棒的到你\(^o^)/~

你好: 是等比数列吧? 由于条件中给出S3,只有三项,所以不用公式,直接展开做比较直观。 解:列式如下。 a1+a1q+a1q^2=9/2...(1) a1q^2=3...(2) 那么,将(2)代入(1)得到:a1+a1q=3;a1(1+q)=3;a1=3/(a+q)...(3) 再把(3)带入到(2)...

S3=a1+a1q+3/2=9/2,a1+a1q=3 a1(1+q)=3,a1=3/(1+q)............................(1) a3=a1q^2=3/2....................................(2) (1)代入(2): 3q^2/(1+q)=3/2,6q^2=3+3q,6q^2-3q-3=0, 2q^2-q-1=0,(q-1)(2q+1)=0,∴q=1或q=-1/2代入(2) a...

q等于负二分之一,a等于六

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