等差数列{an}的公差d∈(0,1),且sin2a3-sin2a7sin(a3+a7)=-1,当n=10时,数列{an}的前n项和Sn取得最小值,则首项a1的取值范围为()A.(-58π,-916π)

题目简介

等差数列{an}的公差d∈(0,1),且sin2a3-sin2a7sin(a3+a7)=-1,当n=10时,数列{an}的前n项和Sn取得最小值,则首项a1的取值范围为()A.(-58π,-916π)

题目详情

等差数列{an}的公差d∈(0,1),且
sin2a3-sin2a7
sin(a3+a7)
=-1
,当n=10时,数列{an}的前n项和Sn取得最小值,则首项a1的取值范围为(  )
A.(-
5
8
π,-
9
16
π)
B.[-
5
8
π,-
9
16
π]
C.(-
5
4
π,-
9
8
π)
D.[-
5
4
π,-
9
8
π]
题型:单选题难度:偏易来源:安徽模拟

答案

∵{an}为等差数列,
sin2a3-sin2a7
sin(a3+a7)
=-1,
1-cos2a3
2
-
1-cos2a7
2
sin(a3+a7)
=-1,
cos2a7-cos2a3
2
=-sin(a3+a7),
由和差化积公式可得:class="stub"1
2
×(-2)sin(a7+a3)•sin(a7-a3)=-sin(a3+a7),
∵sin(a3+a7)≠0,
∴sin(a7-a3)=1,
∴4d=2kπ+class="stub"π
2
∈(0,4)
∴k=0,
∴4d=class="stub"π
2
,d=class="stub"π
8

∵n=10时,数列{an}的前n项和Sn取得最小值,
a10≤0
a11≥0
a1+9×class="stub"π
8
≤0
a1+10×class="stub"π
8
≥0

∴-class="stub"5π
4
≤a1≤-class="stub"9π
8

故选D.

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