It is known that no matter what the value of X is, there is a value of (a + 2b) x + (a-2b) = 3x + 5

It is known that no matter what the value of X is, there is a value of (a + 2b) x + (a-2b) = 3x + 5

A+2B=3
A-2B=5
Add two formulas: 2A = 8, a = 4
Let's take formula one, B = - 1 / 2

In the arithmetic sequence {an}, the known tolerance D is not equal to 0, an is not equal to 0, let equation
A (R) x & # 178; + 2A (R + 1) x + a (R + 2) = 0 (r belongs to N +) is a set of roots of the equation
(1) Prove that the equation has a common root, and find out the common root
(r) And so on are subscripts
What is the common root?
Is Δ = 0?

∵ {an} is an arithmetic sequence, ∵ 2A (R + 1) = a (R) + a (R + 2), that is, a (R) - 2A (R + 1) + a (R + 2) = 0
So when x = - 1, a (R) x ^ 2 + 2A (R + 1) x + a (R + 2)
= a(r)- 2a(r+1)+a(r+2)=0
When r takes different natural numbers, the original equation has a common root-1

If {an} is an arithmetic sequence with tolerance D, then {an + 2A (n + 2)} is an arithmetic sequence with tolerance D

an = a1+(n-1)d
a(n+2)=a1+(n+1)d
an+2a(n+2)=a1+(n-1)d+2a1+2(n+1)d = 3a1+(3n-1)d = 3a1+(3n-3+2)d
=3a1+2d+(n-1)*3d
So the first term is 3A1 + 2D and the tolerance is 3D