Find the tenth term of the geometric sequence with a2=1/2 a5=4?
回答 (3)
2016-08-03 2:27 am
The step is 2.
a2=1/2
a3=1
a4=2
a5=4
a10 = a5 * 2^5 = 4*32 = 128
a2=1/2
a3=1
a4=2
a5=4
a10 = a5 * 2^5 = 4*32 = 128
2016-08-03 2:43 am
any term a(n) = a1 r^(n-1), where a1=1st term and r=geometric ratio
a2 = a1 r^(2-1) = a1 r ....(1)
a5 = a1 r(5-1) = a1 r⁴
a5/a2 = a1 r⁴/a1 r
4/ ½ = r³
8 = r³
r = 2
plug in (1) to get
½ = a1(2)
a1 = ¼
a10 = a1 r^(10-1) = ¼ (2)⁹ = 128
a2 = a1 r^(2-1) = a1 r ....(1)
a5 = a1 r(5-1) = a1 r⁴
a5/a2 = a1 r⁴/a1 r
4/ ½ = r³
8 = r³
r = 2
plug in (1) to get
½ = a1(2)
a1 = ¼
a10 = a1 r^(10-1) = ¼ (2)⁹ = 128
2016-08-03 2:27 am
hint:
rule
a(n) =a1 (r)^(n-1)
find r hten find n(10)
rule
a(n) =a1 (r)^(n-1)
find r hten find n(10)
收錄日期: 2021-04-21 19:28:05
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