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Find the eighth term of the sequence whose first three terms are 4,2,1.?

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  • Find the eighth term of the sequence whose first three terms are 4,2,1.?


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Answer #1 | 17/04 2017 19:11
I'm assuming that is a geometric sequence where each term is 1/2 of the prior term. With only 8 terms, you could do this manually. 4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/32 Or you could use the general formula: a[n] = ar^(n-1) a : first term (4) r : common ratio (½) a[8] = 4*½^7 = 4 * 1/128 = 1/32
Positive: 50 %
Answer #2 | 17/04 2017 19:18
1__2__3_____4_____5____6____7_____8 4__2__1____1/2____1/4__1/8__1/16___1/32 8th term is 1/32
Positive: 0 %
Answer #3 | 17/04 2017 19:18
4, 2, 1, 1/2, ... an = 2^(3 - n) a8 = 1/32
Positive: 0 %
Answer #4 | 17/04 2017 19:12
#1 = a = 4 #2 = ar = 2 #3 = ar^(2) = 1 Hence #3 / #2 = ar^2/ar = 1/2 = r ( By cancelling down). #n = ar^(n-1) #8 = ar^7 = 4 x ( 1/2)^7 = 4/128 = 1/32
Positive: 0 %

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