In the arithmetic sequence {an} with odd terms, s odd denotes the sum of odd terms, s even denotes the sum of even terms, then s odd / s even=

In the arithmetic sequence {an} with odd terms, s odd denotes the sum of odd terms, s even denotes the sum of even terms, then s odd / s even=

Let this arithmetic sequence have 2N-1 terms, then
S odd = n (a1 + a (2n-1)) / 2 = n (2An) / 2 = Nan
S-pairs = (n-1) (A2 + a (2n-2)) / 2 = (n-1) (2An) / 2 = (n-1) an
So s odd / s even = n / (n-1)

The arithmetic sequence {an}, when the number of items is odd or even, find the sum of odd items (s odd) and even items (s even)
There should be a detailed and easy to understand derivation process

If the tolerance is D, then an + 1-an = D
For odd or even terms, if there is an interval between two adjacent terms, then an + 2-An = 2D
S odd = a1 + a3 +... + a (2k-1) (k = 1,2,3...)
=(a1+a(2k-1))*k/2
=(a1+a1+(k-1)*2d)*k/2
=k*a1+k(k-1)d
=k*a1+k²d-kd
S-couple = A2 + A4 +... + a (2k) (k = 1,2,3...)
=(a2+a(2k))*k/2
=(a2+a2+(k-1)*2d)*k/2
=k*a2+k(k-1)d
=k*(a1+d)+k²d-kd
=k*a1+k²d

If the sequence {an} satisfies 3A (n + 1) + an = 0, A2 = - 4 / 3, then the sum of the first 10 terms of {an} is equal to? For detailed explanation, thank you~

3a(n+1)+an=0,
——》a(n+1)/an=-1/3=q,
a2=a1*(-1/3)
——》a1=a2/(-1/3)=(-4/3)/(-1/3)=4,
That is, the sequence {an} is an equal ratio sequence with the first term A1 = 4 and the common ratio q = - 1 / 3,
S10=a1*(1-q^10)/(1-q)
=4*[1-(-1/3)^10]/[1-(-1/3)]
=3-3^(-9).