1.The nth term of a geometric sequence is given by a_n=1/4(4)^n-1 . Write the first five terms of this sequence.
2.The nth term of a a geometric sequence is given by a_n=27(0.1)^n-1 . Write the first five terms of this sequence.
1.
t(1) = (1/4)4^1 - 1 = 1 - 1= 0
t(2) = (1/4)(4^2 - 1 = 4 - 1 = 3
t(3) = (1/4)4^3 - 1 = 4^2 - 1 = 15
t(4) = (1/4)4^4 - 1 = 4^3 - 1 = 63
t(5) = ........ = 255
To find the first five terms of a geometric sequence, you need to substitute the values of n from 1 to 5 into the given formula and calculate the corresponding terms. Let's compute the terms for each sequence:
1. For the sequence a_n = (1/4)(4)^(n-1):
- For n = 1: a_1 = (1/4)(4)^(1-1) = (1/4)(4)^0 = (1/4)(1) = 1/4
- For n = 2: a_2 = (1/4)(4)^(2-1) = (1/4)(4)^1 = (1/4)(4) = 1
- For n = 3: a_3 = (1/4)(4)^(3-1) = (1/4)(4)^2 = (1/4)(16) = 4
- For n = 4: a_4 = (1/4)(4)^(4-1) = (1/4)(4)^3 = (1/4)(64) = 16
- For n = 5: a_5 = (1/4)(4)^(5-1) = (1/4)(4)^4 = (1/4)(256) = 64
Therefore, the first five terms of this sequence are 1/4, 1, 4, 16, and 64.
2. For the sequence a_n = 27(0.1)^(n-1):
- For n = 1: a_1 = 27(0.1)^(1-1) = 27(0.1)^0 = 27(1) = 27
- For n = 2: a_2 = 27(0.1)^(2-1) = 27(0.1)^1 = 27(0.1) = 2.7
- For n = 3: a_3 = 27(0.1)^(3-1) = 27(0.1)^2 = 27(0.01) = 0.27
- For n = 4: a_4 = 27(0.1)^(4-1) = 27(0.1)^3 = 27(0.001) = 0.027
- For n = 5: a_5 = 27(0.1)^(5-1) = 27(0.1)^4 = 27(0.0001) = 0.0027
Hence, the first five terms of this sequence are 27, 2.7, 0.27, 0.027, and 0.0027.