Since a, b, c are in A.P.; b, c, d are in G.P. and

1

c

,

1

d

,

1

e

are in A.P.
Then 2b = a + c ...(i)
Also, c2 = bd
and

2

d

=

1

c

+

1

e

⇒

2ce

d

= c + e
⇒ d =

2ce

c+e

... (iii)
Multiplying (i) and (iii), we get
2bd =

2ce(a+c)

c+e

⇒ bd =

ce(a+c)

c+e

⇒ c2 =

ce(a+c)

c+e

[using (ii)]
⇒ c =

e(a+c)

c+e

⇒ c (c + e) = e (a + c)
⇒ c2 + ce = ea + ce ⇒ c2 = ae
Hence above condition shows that a, c, e are in G.P.