等差数列{an}中,a4=5,且a3,a6,a10成等比数列.(1)求数列{an}的通项公式;(2)写出数列{an}的前10项的和S10.-数学

题目简介

等差数列{an}中,a4=5,且a3,a6,a10成等比数列.(1)求数列{an}的通项公式;(2)写出数列{an}的前10项的和S10.-数学

题目详情

等差数列{an}中,a4=5,且a3,a6,a10成等比数列.
(1)求数列{an}的通项公式;
(2)写出数列{an}的前10项的和S10
题型:解答题难度:中档来源:不详

答案

(1)设数列{an}的公差为d,则
a3=a4-d=5-d,a6=a4+2d=5+2d,a10=a4+6d=5+6d,
由a3,a6,a10成等比数列得a62=a3 a10,
即(5+2d)2=(5-d)( 5+6d),
整理得10d2-5d=0,解得d=0,或d=class="stub"1
2

当d=0时,a4=a1=5,an=5;
d=class="stub"1
2
时,a4=a1+class="stub"3
2
=5

a1=class="stub"7
2
an=class="stub"7
2
+(n-1)×class="stub"1
2
=class="stub"n
2
+3

(2)当d=0时,
S10=10•a4=50.
当d=class="stub"1
2
时,
a1=a4-3d=5-class="stub"3
2
=class="stub"7
2

S10=10×class="stub"7
2
+class="stub"10×9
2
×class="stub"1
2
=class="stub"115
2

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