分析:由题意可得 a1an =3,再由所有项的积为a1•a1q•a1q2 …a1qn-1=243=35 ①,倒序可得 a1qn-1…a1q2•a1q•a1=35 ②,把①②对应项相乘可得
=
=3n=35•35 =310,由此解得 n的值.解答:设等比数列为{an},公比为q,由题意可得 a1a2a3=3,且 an-2an-1an=9,两式相乘可得 a1an =3.
再由所有项的积为a1•a1q•a1q2 …a1qn-1=243=35 ①,
倒序可得 a1qn-1…a1q2•a1q•a1=35 ②,
把①②对应项相乘可得
=
=3n=35•35 =310,解得 n=10,故答案为 10.
点评:本题主要考查等比数列的定义和性质,等比数列的通项公式,属于中档题.