hongy7
幼苗
共回答了28个问题采纳率:89.3% 举报
(1)∵a 1 ,2a 7 ,3a 4 成等差数列,
∴4a 7 =a 1 +3a 4 ,又数列{a n }是首项为a且公比q≠1的等比数列,
∴4aq 6 =a+3aq 3 ,
整理得:4(q 3 ) 2 -3q 3 -1=0,即(4q 3 +1)(q 3 -1)=0,
解得:q 3 =-
1
4 或q 3 =1(舍去),
则q 3 =-
1
4 ;
(2)∵q 3 =-
1
4 ,
∴
S 6
12 S 3 =
a 1 (1- q 6 )
1-q
12 a 1 (1- q 3 )
1-q =
1+ q 3
12 =
1
16 ,
而
S 12 - S 6
S 6 =
S 12
S 6 -1=
a 1 (1- q 12 )
1-q
a 1 (1- q 6 )
1-q -1
= 1+ q 6 -1= q 6 =
1
16 =
S 6
12 S 3 ,
∴S 6 2 =12S 3 •(S 12 -S 6 ),
则12S 3 ,S 6 ,S 12 -S 6 成等比数列.
1年前
9