The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36,

Let the first term of the A.P. be a and the common difference be d.

∴ a = a , b = a + d and c = a + 2d

$a+b+c=18$

$\Rightarrow a+(a+d)+(a+2 d)=18$

$\Rightarrow 3 a+3 d=18$

$\Rightarrow a+d=6$    ....(i)

Now, according to the question, a $+4$, a $+d+4$ and $a+2 d+36$ are in G. P.

$\therefore(\mathrm{a}+\mathrm{d}+4)^{2}=(\mathrm{a}+4)(\mathrm{a}+2 \mathrm{~d}+36)$

$\Rightarrow(6-\mathrm{d}+\mathrm{d}+4)^{2}=(6-\mathrm{d}+4)(6-d+2 \mathrm{~d}+36)$

$\Rightarrow(10)^{2}=(10-\mathrm{d})(42+\mathrm{d})$

$\Rightarrow 100=420+10 d-42 d-d^{2}$

$\Rightarrow d^{2}+32 d-320=0$

$\Rightarrow(d+40)(d-8)=0$

$\Rightarrow d=8,-40$

Now, putting $d=8,-40$ in equation $(\mathrm{i})$, we get, $a=-2,46$, respectively.

For $a=-2$ and $d=8$, we have :

$a=-2, b=6, c=14$

And, for $a=46$ and $d=-40$, we have :

$a=46, b=6, c=-34$