Question ID: #935
In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(\alpha,\beta)$, then $\alpha^{2}+\beta^{2}$ is equal to
Solution:
Let the first three terms of the G.P. be $\frac{A}{r}, A, Ar$.
Given the product is 27:
$$\frac{A}{r} \cdot A \cdot Ar = 27$$
$$A^3 = 27 \Rightarrow A = 3$$
Let $S$ be the sum of the first three terms:
$$S = \frac{3}{r} + 3 + 3r = 3\left( r + \frac{1}{r} + 1 \right)$$
We know that for any real number $r \ne 0$, $r + \frac{1}{r} \ge 2$ (if $r > 0$) or $r + \frac{1}{r} \le -2$ (if $r < 0$).
Case 1: $r > 0$
$$r + \frac{1}{r} + 1 \ge 3$$
$$S = 3\left( r + \frac{1}{r} + 1 \right) \ge 3(3) = 9$$
So $S \in [9, \infty)$.
Case 2: $r < 0$ $$r + \frac{1}{r} + 1 \le -2 + 1 = -1$$ $$S = 3\left( r + \frac{1}{r} + 1 \right) \le 3(-1) = -3$$ So $S \in (-\infty, -3]$.
Combining both cases, the set of all possible values for $S$ is $(-\infty, -3] \cup [9, \infty)$.
This can be written as $\mathbb{R} – (-3, 9)$.
Comparing with $\mathbb{R} – (\alpha, \beta)$, we get:
$$\alpha = -3, \quad \beta = 9$$
We need to find $\alpha^2 + \beta^2$:
$$\alpha^2 + \beta^2 = (-3)^2 + (9)^2 = 9 + 81 = 90$$
Ans. 90
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