Solution:
Consider the first term of an A.P. to be ‘a’ and let its common difference be given by ‘d’.
Therefore, we can write: b = a + d and c = a + 2d.
It is gien that: a + b + c = 18
Or, we can write => 3a + 3d = 18 or a + d = 6.
Therefore, d = 6 – a … (i)
Now, in accordance to the question, we can say that the three terms a + 4, a + d + 4, and a + 2d + 36 are now in GP, that is:
$ \left( a+d+4 \right)/\left( a+4 \right)\text{ }=\text{ }\left( a+2d+36 \right)/\left( a+d+4 \right) $
$ {{\left( a\text{ }+\text{ }d\text{ }+\text{ }4 \right)}^{2}}~=~\left( a\text{ }+\text{ }2d\text{ }+\text{ }36 \right)\left( a\text{ }+\text{ }4 \right) $
$ {{a}^{2}}~+\text{ }{{d}^{2}}~+\text{ }16\text{ }+\text{ }8a\text{ }+\text{ }2ad\text{ }+\text{ }8d\text{ }=\text{ }{{a}^{2}}~+\text{ }4a\text{ }+\text{ }2da\text{ }+\text{ }36a\text{ }+\text{ }144\text{ }+\text{ }8d $
$ {{d}^{2}}-32a-128{{\left( 6-a \right)}^{2}}-32a-128\text{ }=\text{ }0 $
$ 36\text{ }+\text{ }{{a}^{2}}-12a-32a-128\text{ }=\text{ }0 $
$ {{a}^{2}}-44a-92\text{ }=\text{ }0 $
$ {{a}^{2}}-46a\text{ }+\text{ }2a-92\text{ }=\text{ }0 $
$ a\left( a-46 \right)\text{ }+\text{ }2\left( a-46 \right)\text{ }=\text{ }0 $
$ a\text{ }=-2\text{ }or\text{ }a\text{ }=\text{ }46 $
$ d\text{ }=\text{ }6-a $
$ d\text{ }=\text{ }6-\left( -2 \right)\text{ }or\text{ }d\text{ }=\text{ }6-46 $
$ d\text{ }=8\text{ }or-40 $
Then,
in the case when a = -2 and d = 8, the A.P is -2, 6, 14
in the case when a = 46 and d = -40, the A.P is 46, 6, -34
Therefore, the numbers are – 2, 6, 14 or 46, 6, – 34