在正项等比数列an中,若S3=6,a7+a8+a9=24,则S99=?
问题描述：
在正项等比数列an中,若S3=6,a7+a8+a9=24,则S99=?
答
an = a1+(n-1)d
S3= (a1+d)3 = 6 (1)
a7+a8+a9 = 3a1+21d =24 (2)
(2)-(1)
d = 1
a1= 1
S99 = ( a1+49d)99 = 50(99) = 4950an为等比数列an = a1q^(n-1)S3 = a1(1+q+q^2) = 6(1)a7+a8+a9 = a1q^6(1+q+q^2) =24(2)(2)/(1)q^6 = 4q^3 = 2 or -2S99 = S3 + q^3.S3 + q^6.S3+...+ q^96.S3= S3( q^99 -1 )/(q^3-1)S99 = 6(2^33 -1 )S3 = a1(1+q+q^2) = 6 和 a7+a8+a9 = a1q^6(1+q+q^2) =24是怎么得到的？S3 =a1+a2+a3= a1 + a1q +a1q^2=6a7+a8+a9 = a1q^61+a1q^7+a1q^8= a1q^6(1+q+q^2) =24