Question

We have a sequence of 100 positive integers. The average of the first and second numbers is 1. The average of the second and third numbers is 2. The average of the third and fourth numbers is 3. This pattern continues, and the average of the 99th and 100th number is 99. What is the 100th number?

Answer 1

my best guess would be to start out simply writing down what the rules suggest and see if theres a pattern

Answer 2

\[\frac12(a_1+a_2)=1\]
\[\frac12(a_2+a_3)=2\]
\[\frac12(a_3+a_4)=3\]hmm, it looks to me like they are just the numbers .5, 1.5, 2.5 , 3.5 ....

Answer 3

is it 1.5 is integer

Answer 4

"positive integers" .... doh!! sooo close ;)

Answer 5

fine .... since zero is not a positive integer then the only way to get an average of 1 from 2 positive numbers is to add 1 and 1 together

1,1 are the first 2

Answer 6

1+3 averages to 2 so 1,1,3

3+3 averages to 3; 1,1,3,3
1 2 3 4

3+5 averages to 4; 1,1,3,3,5
1 2 3 4

looks to be a doubling of the odds then

Answer 7

\[a_n,a_{n+1}=2n-1~:~n=1,2,3,...\]

Answer 8

(a1+a2)/2 =1
(a1+a2)=2
Now, this is only possible if both a1 and a2 are 1 as both has to be a positive integer.

Answer 9

(a2+a3)/2=2
(1+a3)=4
a3=3

(a3+a4)/2=3
(3+a4)=6
a4=3

(a4+a5)/2=4
(3+a5)=8
a5=5

Answer 10

Looks like a100 =99