geometric sequence
2007-11-12 11:24 pm
The sum to infinite geometric sequence is 16 and the sum to infinity of the squares of its terms is 768/5. Find the common ratio and the fourth term of the sequence.
回答 (1)
2007-11-12 11:33 pm
✔ 最佳答案
S(infinity)=a/(1-r)= 16__________(1)T(1)^2+T(2)^2+...+T(infinity)^2
=(a)^2+(ar)^2+....+[ar^(n-1)]^2+.....
=a^2+a^2r^2+....+a^2r^(2n-2)+.....
=a^2/(1-r^2)
=768/5____________ (2)
(1)^2, we have:
a^2/(1-r)^2=256_______________ (3)
(3)/(2), we have:
(1-r^2)/(1-r)^2=5/3
(1+r)(1-r)/(1-r)^2=5/3
(1+r)/(1-r)=5/3 [where (1-r) not = 0]
3(1+r)=5(1-r)
3+3r=5-5r
8r=2
r=1/4
收錄日期: 2021-04-25 16:53:36
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