Answer :

Let the three numbers be .


According to the question



a + ar + ar2 = 14r…(1)


First two terms are increased by 1, and third decreased by 1



The above sequence is in AP.


We know in AP.


2b = a + c




2ar + 2r = ar2 + a


ar2 – 2ar + a = 2r …(2)


Dividing 1 by 2 we get




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


r = 2 or r = 1/2.


Substituting r = 2 in 2 we get


a(2)2 – 2a(2) + a = 2(2)


4a – 4a + a = 4


a = 4


Substituting r = 1/2 in 2 we get


a(1/2)2 – 2a(1/2) + a = 2(1/2)


a = 4


substituting a and r we get the numbers as 2,4,8.


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