Answer : To find: Three numbers

Given: Three numbers are in G.P. Their sum is 56 Formula used: When a,b,c are in GP, b2 = ac

Let the three numbers in GP be a, ar, ar2 According to condition :-

a + ar + ar2 = 56

a(1 + r + r2) = 56 … (i)

1, 7, 21 be subtracted from them respectively, we obtain the numbers as :- a – 1, ar – 7, ar2 – 21

According to question the above numbers are in AP

⇒ ar – 7 – (a – 1) = ar2 – 21 – (ar – 7)

⇒ ar – 7 – a + 1 = ar2 – 21 – ar + 7

⇒ ar – a – 6 = ar2 – ar – 14

⇒ 8 = ar2 – 2ar + a

⇒ 8 = a(r2 – 2r + 1)

Multiplying the above eqn. with 7

⇒ 56 = 7a(r2 – 2r + 1)

⇒ a(1 + r + r2) = 7a(r2 – 2r + 1)

⇒ 1 + r + r2 = 7r2 – 14r + 7

⇒ 6r2 – 15r + 6 = 0

⇒ 6r2 – 12r – 3r + 6 = 0

⇒ 6r(r – 2) -3 (r – 2) = 0

⇒ (6r – 3) (r – 2) = 0

Or r = 2

Putting          in eqn. (i) a(1 + r + r2) = 56a = 32

The numbers are a, ar, ar2

⇒ 32, 16, 8

a = 8
Putting r = 2 in eqn. (i) a(1 + r + r2) = 56

The numbers are a, ar, ar2

⇒ 8, 16, 32

Ans) We have two sets of triplet as 32, 16, 8 and 8, 16, 32.