Let $$a, ar, ar^2, \ldots$$ be an infinite G.P. If $$\sum_{n=0}^{\infty} ar^n = 57$$ and $$\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$$, then $$a + 18r$$ is equal to

$$S_1 = \frac{a}{1-r} = 57$$. $$S_2 = \frac{a^3}{1-r^3} = 9747$$.

$$\frac{S_2}{S_1^3} = \frac{a^3/(1-r^3)}{a^3/(1-r)^3} = \frac{(1-r)^3}{1-r^3} = \frac{(1-r)^2}{1+r+r^2}$$.

$$\frac{9747}{57^3} = \frac{9747}{185193} = \frac{1}{19}$$.

$$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$$.

$$19(1-2r+r^2) = 1+r+r^2$$. $$19-38r+19r^2 = 1+r+r^2$$. $$18r^2-39r+18 = 0$$. $$6r^2-13r+6 = 0$$.

$$r = \frac{13 \pm \sqrt{169-144}}{12} = \frac{13\pm5}{12}$$. $$r = 3/2$$ or $$r = 2/3$$.

For convergent series: $$r = 2/3$$. $$a = 57(1-2/3) = 19$$.

$$a + 18r = 19 + 12 = 31$$.

The correct answer is Option 3: 31.