There are odd items in the arithmetic sequence {an}, and the sum of odd items is 77, the sum of even items is 66, A1 = 1. Find the number of items and the middle items

There are odd items in the arithmetic sequence {an}, and the sum of odd items is 77, the sum of even items is 66, A1 = 1. Find the number of items and the middle items

Let the number of terms of a sequence be 2n + 1, then s odd = (n + 1) (a1 + A2N + 1) 2 = 77, s even = n (A2 + A2N) 2 = 66 ∥ s odd s even = n + 1n = 7766, ∥ n = 6, the number of terms of a sequence is 13, the middle term is the seventh term, and A7 = 11

It is known that the sequence {an} satisfies A1 = 23, and for any positive integer m, n has am + n = am · an. If the sum of the first n terms of the sequence {an} is Sn, then Sn = n___ ．

∵ am + n = aman holds for any m, n ∵ an = an-1a1 = an-2a12 = A1N = (23) n, so the sequence {an} takes 23 as the first term and 23 as the common ratio. From the first n terms of the sequence and the formula, we can get Sn = 23 [1 - (23) n] 1-23 = 2-2n + 13N, so the answer is: 2-2n + 13N

Q common ratio m.n.k.l ∈ positive integer and M + n = K + L 1. Prove an = am × qn-m 2. Am × an = AK × al
In this paper, we prove that m.n of (1) is preceded by a subscript, followed by a degree of (2) and that klmn is a subscript

In other words, I happened to write this question I'll help you
(1)a(m+1)/a(m)*a(m+2)/a(m+1)*a(m+3)/(am+2)…… an/a(m-1)=q^(n-m)
A (M + 1) a (M + 2) a (M + 3) anything can be eliminated
Then there is an / am = q (n-m)
（2）am=a1*q^(m-1) an=a1*q(n-1) ak=a1*q(k-1) al=a1*q(l-1)
am*an=a1^2*q(m-1+n-1) ak*al=a1^2*q(k-1+l-1)
=a1^2*q(m+n-2) a1^2*(k+l-2)
m+n=k+l
∴a1^2*(m=n-2) =a1^2*(k+l-2)
So am * an = AK * al
I hope I can help you