Q47 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
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 $\_\_\_\_$.
Correct answer: 90
Solution
In the G.P., let the first three terms be $\frac{a}{r}, a, ar$.
From the product condition: $\frac{a}{r} \cdot a \cdot ar = a^3 = 27$, so $a = 3$.
The sum of first three terms is $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + 1 + r)$.
For $r > 0$, by AM-GM: $\frac{1}{r} + r \geq 2$, so $S \geq 9$.
For $r < 0$: $\frac{1}{r} + r \leq -2$.
So $S \leq -3$.
Thus the possible values of $S$ are $(-\infty, -3] \cup [9, \infty) = \mathbb{R} - (-3, 9)$.
Therefore $(a, b) = (-3, 9)$ and $a^2 + b^2 = 9 + 81 = 90$.
From the product condition: $\frac{a}{r} \cdot a \cdot ar = a^3 = 27$, so $a = 3$.
The sum of first three terms is $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + 1 + r)$.
For $r > 0$, by AM-GM: $\frac{1}{r} + r \geq 2$, so $S \geq 9$.
For $r < 0$: $\frac{1}{r} + r \leq -2$.
So $S \leq -3$.
Thus the possible values of $S$ are $(-\infty, -3] \cup [9, \infty) = \mathbb{R} - (-3, 9)$.
Therefore $(a, b) = (-3, 9)$ and $a^2 + b^2 = 9 + 81 = 90$.
Hence, the correct answer is 90.