The sum of three numbers in GP is 56. If 1, 7, 21 be subtracted from them respectively, we obtain the numbers in AP. Find the numbers

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) = 56

a = 32

The numbers are a, ar, ar²

⇒ 32,

⇒ 32, 16, 8

Putting r = 2 in eqn. (i)

a(1 + r + r²) = 56

a = 8

The numbers are a, ar, ar²

⇒ 8,

⇒ 8, 16, 32

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