公差不为零的等差数列{an}中,a3=7,又a2,a4,a9成等比数列. (1)求数列{an}的通项公式. (2)设bn=2an,求数列{bn}的前n项和Sn.
问题描述:
公差不为零的等差数列{an}中,a3=7,又a2,a4,a9成等比数列.
(1)求数列{an}的通项公式.
(2)设bn=2an,求数列{bn}的前n项和Sn.
答
(1)设数列的公差为d,则∵a3=7,又a2,a4,a9成等比数列.∴(7+d)2=(7-d)(7+6d)∴d2=3d∵d≠0∴d=3∴an=7+(n-3)×3=3n-2即an=3n-2;(2)∵bn=2an,∴bn=23n−2∴bn+1bn=23n+123n−2=8∴数列{bn}是等...