There is a column of numbers, the first number is 103, the second number is 51, starting from the third number, each number is the average of the first two numbers, then the number of the 21st number is What is the integral part?

There is a column of numbers, the first number is 103, the second number is 51, starting from the third number, each number is the average of the first two numbers, then the number of the 21st number is What is the integral part?

a1=103
a2=51
a3=77
a4=64
a5=70.5
a6=67.25
a7=68.75
a8=68.0625
The following numbers are between 68.025 and 68.75
So the integral part of the 21st number is 68

The average of data 49, 50, 51, m, n is 50, and the variance is 4. Find the value of M, n

From the average of 49, 50, 51, m, n is 50, we can get 49 + 50 + 51 + m + n = 5 × 50 = 250
m+n=100
The variance is 4
Then 0.2 × ((49-50) & # 178; + (50-50) & # 178; + (51-50) & # 178; + (m-50) & # 178; + (N-50) & # 178;) = 4
The results are as follows
m²+n²-100m-100n+5000=3.6
M = 100-N
(100-N) &# 178; + n & # 178; - 100 (100-N) - 100N = - 4996.4
The solution is N1 = 53, N2 = 47
When n = 53, M = 100-N = 100-53 = 47
When n = 47, M = 100-N = 100-47 = 53
So m = 53, n = 47 or M = 47, n = 53