Question

设Sn是等差数列｛an｝的前n项和，若S3:S6=1:4,则S6:S12=?

Answer 1

S3、S6-S3、S9-S6、S12-S9成等差数列
∴S3+(S9-S6)=2(S6-S3)，(S6-S3)+(S12-S9)=2(S9-S6)
设S3=a，那么S6=4a，∴a+(S9-4a)=2×3a，∴S9=9a
∴3a+(S12-9a)=2×5a，∴S12=16a，∴S6:S12=4a:16a=1:4

望采纳

Answer 2

设首项为a1,公差为d
S3=3a1+(3*2/2)d=3(a1+d)
S6=6a1+(6*5/2)d=6a1+15d=3(2a1+5d)
S3/S6=(a1+d)/(2a1+5d)=1/4
解得d=2a1
S12=12a1+(12*11/2)d=12a1+66d=6(2a1+11d)
∴S6/S12=(2a1+5d)/[2(2a1+11d)]
=12a1/[2(2a1+22a1)]=1/4