In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between
0.0025 and 0.025
0.00025 and 0.0025
0.000025 and 0.00025
0.0000025 and 0.000025
0.00000025 and 0.0000025
题目分析:
等比数列~$a_n$~, ~$a_n$~=0.5∗ ~$a_{n-1}$~; ~$a_10$~ ∈(0.0001,0.001); ~$a_{12}$~=0.5*~$a_{11}$~=0.5*0.5*a~$a_{10}$~=0.25*~$a_{10}$~
所以 ~$a_{10}$~ = 4~$a_{12}$~∈(0.0001,0.001);
~$a_{12}$~ ∈(0.0001/4,0.001/4);
~$a_{12}$~∈(0.000025,0.00025);
综上:答案就是 C.
用带数字的方法。假设在tenth 的位置数字为0。0008,然后1/4后,为0.0002。直接找区间对应。
等比数列:an=a1*q^(n-1),an=q∗ an−1(n>=2,an-1 不等于0,q不等于0),另外注意小数点位置
等比数列: an=a1*q(n-1) ; 顺带记下Sn=a1(1-q的n次方)/1-q
本题:a10=2*2*a12在(0.0001,0.001)之间
所以a12在(0.0001/4,0.001/4)之间
区间