- #1
devanlevin
in a sequence, it is known that the multiplication of the 1st and second digits gives an answer or 4, the multiplication of the 3rd and 4th gives 70
A1*A2=A1*(A1+d)=4
A3*A4=(A1+2d)(A1+3d)=70
A1^2+A1d=4
A1^2+3A1d+2A1d+6d^2=70
A1^2+5A1d+6d^2=70
A1^2=4-A1d
4-A1d+5A1d+6d^2=70
4A1d+6d^2=66
3d^2+2A1d=33
from here i don't know what to do,
A1*A2=A1*(A1+d)=4
A3*A4=(A1+2d)(A1+3d)=70
A1^2+A1d=4
A1^2+3A1d+2A1d+6d^2=70
A1^2+5A1d+6d^2=70
A1^2=4-A1d
4-A1d+5A1d+6d^2=70
4A1d+6d^2=66
3d^2+2A1d=33
from here i don't know what to do,