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

Concept: Let the first three terms of a geometric progression be: \[ \frac{a}{r},\ a,\ ar \] where \( a \neq 0 \) and \( r \neq 0 \). The product and sum of these terms will be used to determine the range of possible values.
Step 1: Use the given product condition \[ \frac{a}{r}\cdot a \cdot ar = a^3 = 27 \] \[ \Rightarrow a=3 \]
Step 2: Write the sum of the first three terms \[ S=\frac{3}{r}+3+3r \] \[ S=3\left(r+\frac{1}{r}+1\right) \]
Step 3: Find the range of the sum For all real \( r\neq 0 \), \[ r+\frac{1}{r}\ge 2 \quad \text{or} \quad r+\frac{1}{r}\le -2 \] Hence, \[ S \ge 3(2+1)=9 \quad \text{or} \quad S \le 3(-2+1)=-3 \] So, the set of all possible values of \( S \) is: \[ (-\infty,-3]\cup[9,\infty) \] \[ \Rightarrow \mathbb{R}-(a,b)=\mathbb{R}-(-3,9) \] Thus, \[ a=-3,\quad b=9 \]
Step 4: Compute \( a^2+b^2 \) \[ a^2+b^2=(-3)^2+9^2=9+81=90 \] Final Answer: \[ \boxed{90} \]