If a,b,c,d are in G.P, then show that a-b, b-c, and c-d are in G.P.?

Question

please show steps

Answer 1

In a geometric sequence each term is found by multiplying the previous term by a constant.

The sequence is a, ar, ar², ar³,....

where r is common ratio.

a, b, c, d are in G. P

b/a = r then b = ar ---> (1)

c/b = r then c = br ---> (2)

d/c = r then  d = cr ---> (3)

Now the series (a - b), (b - c), (c - d)

a - b = a - ar

a - b = a(1 - r) ---> (4)

b - c = b - br 

= b(1 - r)

b - c = ar(1 - r) ---> (5)

c - d = c -  cr

= c(1 - r)

= br (1 - r)

= arr(1 - r)

c - d = ar²(1 - r) ---> (6)

The series can be written as a(1 - r), ar(1 - r), ar²(1 - r)

In this case common ratio = r.

Therefore, (a - b), (b - c), (c - d) are in GP.

Answer 2

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Answer 3

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