If the arithmetic sequence 5,8,11,... And 3,7,11,... Have 100 terms, then the number of terms of their same terms is___ . But I don't understand. Please tell me, ∵3(4n-1)+2≤a100=302 ∴n≤25.

If the arithmetic sequence 5,8,11,... And 3,7,11,... Have 100 terms, then the number of terms of their same terms is___ . But I don't understand. Please tell me, ∵3(4n-1)+2≤a100=302 ∴n≤25.

The general formula of 5,8,11,... Is an = 3n-1, n ≤ 100
The general formula of 3,7,11,... Is BK = 4k-1, K ≤ 100
Let 3n-1 = 4k-1
So n = (4K) / 3
k=3、6、9、12、…… seventy-five
So there are 75 △ 3, that is, 25 items are the same

It is known that two arithmetic sequences 5,8,11. And 3,7,11. Have 100 terms. How many terms do they have in common?
One kind of solution: the general term formula of two sequences is 3N + 2 = 4m-1, then n = 4 / 3m-1 can be obtained. Let m = 3R (R is a positive integer), then n = 4r-1 can be obtained
So 1 ≤ 3R ≤ 100, 1 ≤ 4r-1 ≤ 100, the solution is 1 ≤ R ≤ 25, so 25 terms are the same
Why let m = 3R?

Because n is a positive integer, so 4 / 3m-1 is also a positive integer, then M divided by 3 must be a positive integer, so let m = 3R, R be a positive integer (that is, M can be divided by 3)

Two arithmetic sequences 5,8,11... And 3,7,11... Have 100 items. How many items are the same?
Please write down the detailed process and results

The general term of the first sequence an = 5 + 3 (n-1) = 3N + 2. (n = 1,2,3,..., 100)
The general term of the second sequence BM = 3 + 4 (m-1) = 4m-1. (M = 1,2,3,..., 100)
The same term in two sequences satisfies 3N + 2 = 4m-1, that is, 3 (n + 1) = 4m
When n = 4k-1, 3 (n + 1) = 3 (4k-1 + 1) = 12K = 4m, M = 3K (k = 1,2,3,..., 25)
That is, when k is the same, a (4k-1) in {an} is the same as B (3K) in {BM}
That is to say, the third, sixth, ninth, twelfth, fifteenth, eighteenth, seventy-five terms of {BM} are equal to the third, seventh, eleventh, fifteenth, nineteenth, twenty-three, ninety-nine terms of {an}