,求S5.在线课程解:数列{an}是等比数列,Sn是它的前n项和,若a2•a3=2a1=a1•a4,可得a4=2.再由a4与2a7的等差中项为
,可得a4 +2a7 =
,故有a7 =
.∴q3=
=
,∴q=
,∴a1=16.∴s5=
=31.分析:由a2•a3=2a1=a1•a4,可得a4=2,再由a4与2a7的等差中项为
,得a4 +2a7 =
,故有a7 =
.求出首项和公比,再利用等比数列的前n项和公式求出s5.点评:本题主要考查等差数列的定义和性质,等比数列的通项公式,等比数列的前n项和公式,属于中档题.