依题意,得a-d+a+a+d=15,解得a=5
所以{bn}中的依次为7-d,10,18+d
依题意,有(7-d)(18+d)=100,解得d=2或d=-13(舍去)
故{bn}的第3项为5,公比为2
由b3=b1•22,即5=4b1,解得b1=
5 |
4 |
所以{bn}是以
5 |
4 |
5 |
4 |
(II)数列{bn}的前和Sn=
| ||
1− 2 |
5 |
4 |
5 |
4 |
即Sn+
5 |
4 |
5•2n |
4 |
5 |
4 |
5 |
2 |
Sn+1+
| ||
Sn+
|
5•2n−1 |
5•2n−2 |
因此{Sn+
5 |
4 |
5 |
2 |
5 |
4 |
5 |
4 |
5 |
4 |
5 |
4 |
| ||
1− 2 |
5 |
4 |
5 |
4 |
5 |
4 |
5•2n |
4 |
5 |
4 |
5 |
2 |
Sn+1+
| ||
Sn+
|
5•2n−1 |
5•2n−2 |
5 |
4 |
5 |
2 |