Question

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 \( \mathbb{R} - (a,b) \), then \( a^2+b^2 \) is equal to:

Hint:

For expressions of the form \( r+\frac{1}{r} \), always use the inequality \( r+\frac{1}{r}\ge 2 \) or \( \le -2 \) to determine the range.

Answer 1

Correct Answer: 90

Step 1: Understanding the Concept:
Let the first three terms of the Geometric Progression (G.P.) be $\frac{x}{r}$, $x$, and $xr$.
We are given the product of these terms and need to find the range of their sum.
Step 2: Key Formula or Approach:
Product of terms $= \left(\frac{x}{r}\right)(x)(xr) = x^3$.
Sum of terms $S = x\left(\frac{1}{r} + 1 + r\right)$.
For real $r$, the range of $r + \frac{1}{r}$ is $(-\infty, -2] \cup [2, \infty)$.
Step 3: Detailed Explanation:
Given: $x^3 = 27 \implies x = 3$.
The sum $S = 3\left(1 + r + \frac{1}{r}\right)$.
Let $y = r + \frac{1}{r}$. Since $r$ is a real common ratio, $y \in (-\infty, -2] \cup [2, \infty)$.
Case 1: $y \geq 2$.
$S = 3(1 + y) \geq 3(1 + 2) = 9$.
Case 2: $y \leq -2$.
$S = 3(1 + y) \leq 3(1 - 2) = -3$.
Thus, the set of all possible values for the sum $S$ is $(-\infty, -3] \cup [9, \infty)$.
Comparing this with $\mathbb{R} - (a, b)$, we find the excluded open interval is $(-3, 9)$.
Therefore, $a = -3$ and $b = 9$.
The value of $a^2 + b^2 = (-3)^2 + (9)^2 = 9 + 81 = 90$.
Step 4: Final Answer:
The value of $a^2 + b^2$ is 90.