In the arithmetic sequence {an}, A1 = 25 and S9 = S17 (1) find the general term formula of the sequence {an}, and (2) find the number of terms before the sequence and the maximum

In the arithmetic sequence {an}, A1 = 25 and S9 = S17 (1) find the general term formula of the sequence {an}, and (2) find the number of terms before the sequence and the maximum

S9=9x25+9x8d/2=S17=17x25+17x16d/2
d=-2
So an = 27-2n
From an > = 0, 27-2n > = 0, n is obtained

In the arithmetic sequence {an}, A1 = 25, S9 = S17, then______ Item and maximum

∵ A1 = 25, S9 = S17, ∵ 9a1 + 9 × 82d = 17a1 + 17 × 162d, the solution is d = - 2. ∵ Sn = 25N + × n (n − 1) 2 (- 2) = - N2 + 26n = - (N-13) 2 + 169

In the arithmetic sequence {an}, A1 = 25, S17 = S9, ask how many items in the sequence have the largest sum, and find out the maximum

Solution 1: ∵ A1 = 25, S17 = S9, ∵ 17a1 + 17 × 162d = 9a1 + 9 × 82d, the solution is d = - 2. ∵ Sn = 25N + n (n − 1) 2 × (- 2) = - N2 + 26n = - (N-13) 2 + 169

In the arithmetic sequence {an}, A1 = 25, S17 = S9, how many terms and maximum of this sequence? What is the maximum value?
It's S17 = A9. It's S17 = S9. Wrong number

From S17 = S9, we can know that A10 + a11 + A12 + A13 + A14 + A15 + a16 + A17 = 0, we can know that the tolerance D is less than 0, and A13 + A14 = 0, so the 13 items and the maximum (A14 is less than 0) of this sequence and A13 + A14 = 0, that is 12D + 25 + 13D + 25 = 0, d = - 2, the 13th item is 25-24 = 1s13 = (25 + 1) * 13 / 2 = 169

In the arithmetic sequence {a (n)}, we know the general term formula of A1 = 25, S9 = S17, an and the maximum sum of the first several terms of the sequence, and find the maximum

A1 = 25, substituting into the sum formula Sn = n * a1 + n (n-1) d / 2, it is obtained that:
Sn=25n+n(n-1)d/2
From S17 = S9, d = - 2,
Sn= -n^2+26n=-(n-13)^2+169
When n = 13, Sn has a maximum value of 169

It is known that the sequence {an} is an arithmetic sequence, Sn is the sum of its first n terms, a1 + A5 = 6, S9 = 63. (1) find the general term formula an and the first n terms and Sn of the sequence {an}; (2) the sequence {BN} satisfies: for ∀ n ∈ n *, BN = 2An, find the first n terms and TN of the sequence {an · BN}

(1) ∵ S9 = 63, ∵ 9A5 = 63, the solution is A5 = 7. ∵ a1 + A5 = 6, ∵ A1 = - 1, ∵ d = A5 − A14 = 2, ∵ an = 2n-3, Sn = N2 − 2n. (2) ∵ an = 2n-3, BN = 2An,