Question

Product of first 3 terms of a G.P. is 27 and sum is \(R - \{a,b\}\), then \(a^2 + b^2\) is equal to :

• A. 90
• B. 81
• C. 9
• D. 18

Hint:

For any real \(r \neq 0\), the expression \(r + \frac{1}{r}\) always lies in \((-\infty, -2] \cup [2, \infty)\). Multiplying by a constant and adding a term preserves this "gap" in the range.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The sum of three terms in a Geometric Progression (G.P.) has a specific range of values for real common ratios. The values excluded from the range determine the parameters \(a\) and \(b\).
Step 2: Key Formula or Approach:
Let the terms be \(\frac{A}{r}, A, Ar\).
Product = \(A^3\).
Sum \(S = A(\frac{1}{r} + 1 + r)\).
Step 3: Detailed Explanation:
Given Product = 27, so \(A^3 = 27 \implies A = 3\).
Sum \(S = 3(\frac{r^2 + r + 1}{r})\).
Let \(y = \frac{r^2 + r + 1}{r} \implies r^2 + (1-y)r + 1 = 0\).
For \(r\) to be real, the discriminant \(D \ge 0\):
\[ (1-y)^2 - 4(1)(1) \ge 0 \implies (y-1)^2 \ge 4 \implies |y-1| \ge 2 \] This gives \(y \ge 3\) or \(y \le -1\).
Substituting back for \(S = 3y\):
\[ S \ge 9 \text{ or } S \le -3 \] The possible values for the sum are \((-\infty, -3] \cup [9, \infty)\).
The set of values that the sum *cannot* take is the interval \((-3, 9)\).
Identifying the boundaries as \(a\) and \(b\), we have \(a = -3\) and \(b = 9\).
Thus, \(a^2 + b^2 = (-3)^2 + 9^2 = 9 + 81 = 90\).
Step 4: Final Answer:
The value of \(a^2 + b^2\) is 90.