Question

设Sn为等差数列{an}的前n项和,若S3=3,S6=24,则a9多少?

Answer 1

设首项为a1,公差为d

s3=a1+a2+a3

=a1+(a1+d)+(a1+2d)

=3a1+3d

s6=s3+a4+a5+a6

=3a1+3d+(a1+3d)+(a1+4d)+(a1+5d)

=6a1+15d

3a1+3d=3

6a1+15d=24

a1=-1,d=2

a9=a1+8d=15

设首项为a1,公比为q

a1a2a3=a1(a1q)(a1q^2)=a1^3q^3

a7a8a9=(a1q^6)(a1q^7)(a1q^8)=a1^3q^21

(a1a2a3)(a7a8a9)=a1^6q^24=5*10=50

a1^3q^12=√50

a4a5a6=(a1q^3)(a1q^4)(a1q^5)=a1^3q^12=√50