Find three consecutive integers such that the sum of the first and twice the third is one more than three times the second.

So we have A, B and C
B= A+1
C= A + 2
A + 2C = 3B + 1 
Now substitute for A

A + 2(A+2) = 3(A+1) +1   now distribute  A + 2A + 4= 3A + 3 + 1 now simplify

3A + 4 =  3A  + 4   since the values are equal you need to only find any three   consecutive integers

Geometric pattern of 2 i’m guessing. not really sure about what you’re asking but i hope this !

Its d so thats how it would would back from 1.0 form three four two