Let a, ar, ar2 be three numbers in G.P. whose sum is 56, then
a + ar + ar2 = 56 ⇒ a(1 + r + r2) = 56 .....(i)
We are given, if 1, 7, 21 are subtracted from a, ar, ar2 respectively, then
they form an A.P. i.e., a – 1, ar – 7, ar2 – 21, ...... form an A.P. Then
2(ar – 7) = (ar2 – 21) + (a – 1) ⇒ 2ar – 14 = ar2 + a – 22
⇒ ar2 + a – 2ar = – 14 + 22
⇒ a(r2 – 2r + 1) = 8 .... (ii)
Dividing (i) by (ii), we get

a(1+r+r2)

a(r2−2r+1)

=

56

8

⇒

1+r+r2

1−2r+r2

= 7 ⇒ 1 + r + r2 = 7 - 14r + 7r2
⇒ 6r2 - 15r + 6 = 0 ⇒ 2r2 - 5r + 2 = 0
⇒ (2r - 1) (r - 2) = 0 ⇒ r =

1

2

, 2
If r =

1

2

, then a (1 + r + r2) = 56
⇒ a(1+

1

2

+

1

4

) = 56 ⇒ a(

4+2+1

4

) = 56 ⇒ a =

56×4

7

= 32
Then, the numbers are 32, 16, 8
If r = 2, then a(1 + r + r2) = 56 ⇒ a(1 + 2 + 4) = 56 ⇒ a =

56

7

= 8
Then, the numbers are 8, 16, 32.
Hence, required numbers are 8, 16, 32.