JEE Mathematics 2026 Question with Solution

Given: The product of the first three terms of a G.P. is $27$.

Find: The value of $a^2+b^2$ if the set of all possible sums of the first three terms is $\mathbb{R}-(a,b)$.

Let the first three terms of the geometric progression be

$$\frac{a}{r},\ a,\ ar$$

where $a \neq 0$ and $r \neq 0$.

Using the given product condition,

$$\frac{a}{r}\cdot a \cdot ar = a^3 = 27$$

So,

$$a = 3$$

Now the sum of the first three terms is

$$S = \frac{3}{r} + 3 + 3r$$

Thus,

$$S = 3\left(r + \frac{1}{r} + 1\right)$$

For all real $r \neq 0$,

$$r + \frac{1}{r} \ge 2 \quad \text{or} \quad r + \frac{1}{r} \le -2$$

Therefore,

$$S \ge 3(2+1) = 9$$

or

$$S \le 3(-2+1) = -3$$

Hence, the set of all possible values of $S$ is

$$(-\infty,-3] \cup [9,\infty)$$

So,

$$\mathbb{R}-(a,b) = \mathbb{R}-(-3,9)$$

which gives

$$a = -3, \quad b = 9$$

Now,

$$a^2+b^2 = (-3)^2 + 9^2 = 9 + 81 = 90$$

Therefore, the required value is $90$.](streamdown:incomplete-link)

Given: The first three terms form a G.P. and their product is $27$.

Find: The endpoints $a$ and $b$ of the excluded interval, then compute $a^2+b^2$.

Write the three terms as

$$\frac{a}{r},\ a,\ ar$$

Their product is

$$\frac{a}{r}\cdot a \cdot ar = a^3$$

Since this equals $27$,

$$a^3 = 27 \Rightarrow a = 3$$

So the sum becomes

$$S = \frac{3}{r}+3+3r = 3\left(r+\frac{1}{r}+1\right)$$

Now use the standard range result for real nonzero $r$:

$$r+\frac{1}{r} \in (-\infty,-2] \cup [2,\infty)$$

Adding $1$ gives

$$r+\frac{1}{r}+1 \in (-\infty,-1] \cup [3,\infty)$$

Multiplying by $3$ gives

$$S \in (-\infty,-3] \cup [9,\infty)$$

Thus the missing open interval is

$$(a,b)=(-3,9)$$

Therefore,

$$a^2+b^2 = (-3)^2 + 9^2 = 90$$

Therefore, the required numerical answer is $90$.](streamdown:incomplete-link)