G.P. Terms Analysis

Let the first three terms of the Geometric Progression (G.P.) be represented as $\frac{x}{r}$, $x$, and $xr$. This representation ensures their product maintains a simpler form.

Product of G.P. Terms

The problem states the product of the first three terms is 27:

$ \left( \frac{x}{r} \right) \times x \times (xr) = 27 $

Simplifying this equation gives:

$ x^3 = 27 $

Taking the cube root, we find the middle term:

$ x = 3 $

So, the first three terms are $\frac{3}{r}$, 3, and $3r$. Note that $r \neq 0$. The common ratio $r$ can be any real number except 0.

Sum of G.P. Terms

The sum ($S$) of these three terms is:

$ S = \frac{3}{r} + 3 + 3r $

Factor out 3:

$ S = 3 \left( \frac{1}{r} + 1 + r \right) $

Let $y = r + \frac{1}{r}$. The sum becomes:

$ S = 3(1 + y) $

Possible Values for Sum ($S$)

We need to determine the range of possible values for $y = r + \frac{1}{r}$.

• Case 1: $r > 0$

  Using the AM-GM inequality ($r + \frac{1}{r} \ge 2\sqrt{r \times \frac{1}{r}}$), we get $y \ge 2$. In this case, $1+y \ge 3$, so $S = 3(1+y) \ge 9$. The sum range is $[9, \infty)$.

• Case 2: $r < 0$

  Let $r = -k$, where $k > 0$. Then $y = -k + \frac{1}{-k} = -\left( k + \frac{1}{k} \right)$. Since $k > 0$, $k + \frac{1}{k} \ge 2$. Therefore, $y \le -2$. In this case, $1+y \le -1$, so $S = 3(1+y) \le -3$. The sum range is $(-\infty, -3]$.

Combining both cases, the set of all possible values for the sum $S$ is $(-\infty, -3] \cup [9, \infty)$.

Identifying Excluded Sum Values

The problem states the set of all possible sums is $\mathbb{R} - (a, b)$. This means the interval $(a, b)$ contains all the real numbers that are *not* possible values for the sum $S$. Comparing this with our result $(-\infty, -3] \cup [9, \infty)$, the excluded interval is $(-3, 9)$. Therefore, we identify $a = -3$ and $b = 9$.

Calculating $a^2 + b^2$

Finally, we calculate $a^2 + b^2$ using the values $a = -3$ and $b = 9$:

$ a^2 + b^2 = (-3)^2 + (9)^2 $

$ a^2 + b^2 = 9 + 81 $

$ a^2 + b^2 = 90 $