a13=a1+12d
42=6+12d
8=12d
d=2/3
an=a1+d
an=6+2/3
√(x−1)+2=a
√(x−1)=a -2
ОДЗ х-1 ≥0. х≥1
а-2≥0. а≥2
a ∈ [ 2 ; + ∞ )