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.

s3=a3(1/q²+1/q+1)=3/2(1+q+q²)/q²=9/2 得:1+q+q²=3q² 2q²-q-1=0 (2q+1)(q-1)=0 q=-1/2, 1 当q=-1/2时, a1=a3/q²=3/2/(-1/2)²=6 当q=1时, a1=a3=3/2

解: S3=9/2 a3/q²+a3/q+a3=9/2 a3=3/2代入,整理,得 2q²-q-1=0 (q-1)(2q+1)=0 q=1或q=-½ q=1时,a1=a3=3/2 q=-½时,a1=a3/q²=(3/2)/(-½)²=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 q = 1 或 q = -1/2 当 q = 1 时 ,a1 = 3/2 当 q = -1/2 时 ,a1 = 6

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 = a3 + a1 +a2 = a3/q^2 + a3/q + a3 所以 9/2 = 3/2(1+ 1/q^2+1/q) 解得q=1或q=-1/2 q=1时,a1=3/2;q=-1/2时,a1=6

a3=a1q^2=3/2, S3=a1+a1q+a1q^2=9/2, 相除得(1+q+q^2)/q^2=3, ∴2q^2-q-1=0, ∴q=1,或q=-1/2.

解:3/2=a3=q^2*a1 9/2=S3=a1+a2+a3=a1+q*a1+3/2 故3q^2=(9/2-3/2)*q^2=(a1+q*a1+3/2-3/2)*q^2 =a1*q^2+a1*q^2*q =3/2+(3/2)q 解之得 q=1或(-1/2) 若q=1则a1=a3/(q^2)=a3=3/2 若q=(-1/2)则a1=a3/(q^2)=a3=6

a1+a2q+a3q=9/2

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