In the arithmetic sequence {an}, a2-a1 = 8, and A4 is the proportional median of A2 and A3

In the arithmetic sequence {an}, a2-a1 = 8, and A4 is the proportional median of A2 and A3

d=a2-a1=8
a4^2=a2*a3
(a1+3d)^2=(a1+d)(a1+2d)
a1^2+6d*a1+9d^2=a1^2+3d*a1+2d^2
a1=-7d/3=-56/3
an=a1+(n-1)d
=-56/3+8(n-1)
=8n-80/3

Given f (x + 1) = (x ^ 2) - 4, in the arithmetic sequence {an}, A1 = f (x-1), A2 = - (3 / 2), A3 = f (x), find the general formula of {an}

f(x+1)=(x^2)-4=(x+1+1)(x+1-3),so f(x)=(x+1)(x-3),f(x-1)=x(x-4)because a3+a1=2a2,so (x+1)(x-3)+x(x-4)=-3 so x=0 or 3 when x=0,-3,-(3/2),0 so an=-3+(3/2)(n-1) when x=3,0,-(3/2),-3 so an=-(3/2)(n-1)

In the arithmetic sequence {an}, it is known that A1 = 1, A2 = 3 of X + 1, A3 = 9 of X-2 to find A2, A3, A4

A1 = 1, A2 = 3 ^ x + 1, A3 = 9 ^ X-2 are arithmetic sequences, then
A3-a2 = a2-a1, that is, A3 + A1 = 2 * A2
9^x-2+1=2*（3^x+1）
(3^x)^2 -2* 3^x -3=0
3 ^ x = 3 or 3 ^ x = - 1, the latter is inconsistent
So x = 1