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Two arithmetic sequences 5,8,11. And 3,7,11. Have 100 items. How many items do they have in common?
It is easy to calculate the general formula of these two arithmetic sequences
an=5+3（n-1）=3n+2,1≤n≤100
am=3+4（m-1）=4m-1,1≤m≤100
3N + 2 = 4m-1, that is, 3 (n + 1) = 4m
So n + 1 must be a multiple of 4, and M must be a multiple of 3
100/4=25
100 / 3 = 33 + 1
25＜33
So there are 25 items in common
If two arithmetic sequences 5,8,11 And 3,7,11 How many of them are the same
25
The last term of the former sequence: 5 + (100-1) * 3 = 302; the last term of the latter sequence: 3 + (100-1) * 4 = 399
The same number in them can not exceed 302. So the same number should be 302 / (3 * 4) = 25... 2, that is 25
There is an arithmetic sequence; 500495490485,..... What is the 50th term of this arithmetic sequence?
Tolerance d = 5
Therefore, item 50 = 500-5 * (50-1) = 255
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