Answer: 2048, 512, 128, 32, 8
Step-by-step explanation:
Recognize 2048 is 2^11, as all hackers know
1024 is 2^10 :-). Also 8 is 2^3.
So this problem is exactly similar to filling in the geometric sequence starting at 10^11 = 100,000,000,000, with three intermediate values ending with 10^3 = 1000.
(100,000,000,000) = 10^11
1,000,000,000 = 10^9
10,000,000 = 10^7
100,000, = 10^5
(1,000) = 10^3
Or, it is exactly similar to filling in the arithmetic series 11, a, b, c, 3. That one is (11), 9, 7, 5, (3), where we subtract 2 each time.
[tex]A_0 = 11[/tex]
[tex]A_{n+1} = A_{n} - 2[/tex]
The answer is (2048), 512, 128, 32, (8) where we divide by 2^2 each time. Or (2^11), 2^9, 2^7, 2^5, (2^3).
[tex]G_0 = 2^{11}=512[/tex]
[tex]G_{n+1} = G_{n} / 2^2[/tex]
