Answer (A)
Using eucledian algorithm i.e.
If a = bq + r, then GCD(a, b) = GCD(b, r).
using this in question
a= $k_{1}b + r_{1}$ => GCD(a,b)= GCD(b,$r_{1}$)
b= $k_{2},$r_{1}$ + r_{2}$ => GCD(b,r1)= GCD($r_{1}$,$r_{2}$)
hence GCD(a,b) = GCD(b,r1)= GCD($r_{1}$,$r_{2}$)
Prove of eucledian algorithm
If a = bq + r, then an integer d is a common divisor of a and b if, and only if, d is a common divisor of b and r.
Let d be a common divisor of a and b. Then d divides a and d divides b. Thus d divides (a − bq), which means d disvides r, since r = a − bq. Thus d is a common divisor of b and r.
a = bq + r, then GCD(a, b) = GCD(b, r)