Let the sum of odd and even terms be 44 and 33, respectively

Let the sum of odd and even terms be 44 and 33, respectively

Let {an} be 2n + 1, s odd = a1 + a3 + +A2N + 1 = (n + 1) (a1 + A2N + 1) 2 = (n + 1) an + 1, s-even = A2 + A4 + A6 + +A2N = n (A2 + A2N) 2 = Nan + 1, s odd s even = n + 1n = 4433, the solution is n = 3, 2n + 1 = 7, because s odd s even = an + 1 = a, so A4 = s odd s even = 44-33 = 11, so the middle term is 11

Given that the number of terms n of arithmetic sequence {a} is odd, and the sum of odd terms s = 44, the sum of even terms t = 33, find the number of terms n

Sum of odd terms s = (a1 + an) / 2 × [(n-1) / 2 + 1] = 44
The sum of even terms t = (A2 + an-1) / 2 × (n-1) / 2 = 33
Because it is an arithmetic sequence, a1 + an = A2 + an-1
(1) formula divided by (2) formula: (n + 1) / (n-1) = 4 / 3
n=7

If the sum of the first n terms of the equal difference {an} is 377, the number of terms n is odd, and in the sum of the first n terms, the ratio of the sum of the odd and even terms is 7:6, find the middle term

Sum of odd items = number of odd items × middle items
Sum of even items = number of even items × middle items
Therefore, the number of odd items: the number of even items = the sum of odd items: the sum of even items = 7:6
Therefore, there are seven odd terms, six even terms, and the middle term = 377 / 13 = 29