Answer :

Let the three numbers be


According to the question


…(1)


a + ar + ar2 = 56r



…(2)


Subtracting 1,7,21 we get,



The above numbers are in AP


If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c



2ar – 14r = a – r + ar2 – 21r


ar2 – 8r + a – 2ar = 0


a(r2 – 2r + 1) = 8r


From (2) we know the value of a



56(r2 – 2r + 1) = 8(1 + r + r2)


7(r2 – 2r + 1) = (1 + r + r2)


7r2 – 14r + 7 = 1 + r + r2


6r2 – 15r + 6 = 0


6r2 – 12r – 3r + 6 = 0


6r(r – 2) – 3(r – 2) = 0


r = 2 or r = 3/6 = 1/2


When r = 2 a = 16 {using equation 1)


r = 1/2 a = 16


the three numbers are (a/r, a, ar) = (8,16,32)


Or numbers are – (32,16,8)


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