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.


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