數(shù)列{an}中,an=32,sn=63,
(1)若數(shù)列{an}為公差為11的等差數(shù)列,求a1;
(2)若數(shù)列{an}為以a1=1為首項(xiàng)的等比數(shù)列,求數(shù)列{am2}的前m項(xiàng)和sm′.
分析:(1)由數(shù)列為等差數(shù)列,根據(jù)條件,用首項(xiàng)和公差分別表示通項(xiàng)和前n項(xiàng)和建立方程組求解.
(2)由數(shù)列為等比數(shù)列,根據(jù)條件,用首項(xiàng)和公比分別表示通項(xiàng)和前n項(xiàng)和建立方程組求解.
解答:解:(1)∵
=sn=63,
a
1+(n-1)11=a
n=32
解得 a
1=10.
(2)
a1×qn-1=32,=63 解得:q=2 n=6
∴所以{a
n2}是首項(xiàng)為1,公比為4的等比數(shù)列
∴S
m=
= 點(diǎn)評(píng):本題主要考查等差數(shù)列和等比數(shù)列的通項(xiàng)公式和前n項(xiàng)和公式的應(yīng)用,這里用的首項(xiàng)和公差,公比,應(yīng)用了方程思想.