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.