geometric sequence

Given that four distinct real number a , b , c and d are the consecutive terms of a geometric sequence and r is the common ratio of the sequence.
(a)Express b ,c and d in terms of a and r.
b = ar
c = ar2
d = ar3
(b)Hence show that a-b , b-c , c-d is also a geometric sequence. What is the common ratio of this germetric sequence.
a – b = a – ar = a(1 – r)
b – c = ar – ar2 = a(1 – r)r
c – d = ar2 – ar3 = a(1 – r)r2
因為
(c – d)/(b – c) = a(1 – r)r2 / a(1 – r)r = r
(b – c)/(b – a) = a(1 – r)r / a(1 – r) = r
所以
(c – d)/(b – c) = (b – c)/(b – a)
因此是等比級數。
它的首項為 a(1 – r)
而公比為 r

answer:(a)b = ar , c = ar2 , d = ar3 (b)r

a)
since r is the common ratio of the geometric sequence a, b, c ,d
we get b/a = c/b = d/c = r
so that b = ar
and c = br = ar^2
and d = cr = ar^3
b)
a - b = a( 1 - r)
b - c = ar( 1 - r)
c - d = ar^2 (1 - r)
so that, (b - c ) / (a - b) = (c - d) / (b - c) = r
therefore, a-b , b-c , c-d is also a geometric sequence with common ratio "r"