Question:

Step 1: Use the G.P. condition. For $a$, $b$, $b-2$ to be in G.P., $b^2 = a(b-2)$. $$b^2 - ab + 2a = 0$$ Step 2: Solve for $b$ using the quadratic formula. $$b = \frac{a \pm \sqrt{a^2 - 8a}}{2}$$ For real $b$: $a^2 - 8a \geq 0 \Rightarrow a(a-8) \geq 0 \Rightarrow a \leq 0$ or $a \geq 8$. Since $a \in N$, we have $a \geq 8$. Step 3: For $a \in [1,8]$, only $a = 8$ is valid (natural number with $a \geq 8$). When $a = 8$: $b = \frac{8 \pm 0}{2} = 4$. So the G.P. is $8, 4, 2, \ldots$ with $r = \frac{4}{8} = \frac{1}{2}$. Option (1) is correct. Step 4: Compute $S_\infty$ for $a = 8$, $r = \frac{1}{2}$. $$S_\infty = \frac{a}{1-r} = \frac{8}{1 - \frac{1}{2}} = 16$$ Option (2) is correct. Step 5: For $a \in (8, 11]$, i.e., $a = 9, 10, 11$ (natural numbers). For $a = 9$: $b = \frac{9 \pm \sqrt{81-72}}{2} = \frac{9 \pm 3}{2}$, so $b = 6$ or $b = 3$. - If $b = 6$: G.P. is $9, 6, 4, \ldots$, $r = \frac{2}{3}$, $S_\infty = \frac{9}{1-2/3} = 27$. So $S_\infty = 27$ is achievable. Option (3) is correct. Step 6: Check option (4). For $a \in (1,8]$, only $a=8$ is valid giving $S_\infty = 16 \neq \frac{27}{2}$. Option (4) is incorrect.