JEE Mathematics 2026 Question with Solution

Given: $a, b, c$ are in A.P., $a^2, 2b^2, c^2$ are in G.P., $a < b < c$, and $a+b+c = 1$.

Find: The value of $9(a^2+b^2+c^2)$.

From the solution, write the A.P. terms as

$$a = b-d, \quad c = b+d$$

Using the G.P. condition,

$$(2b^2)^2 = a^2c^2$$

so

$$4b^4 = (b-d)^2(b+d)^2 = (b^2-d^2)^2$$

The extracted working then takes square root and gets

$$2b^2 = b^2 - d^2$$

which gives

$$d^2 = -b^2$$

The working then states the admissible case as

$$d^2 = b^2 \Rightarrow d = b$$

Using the sum condition,

$$a+b+c = (b-d)+b+(b+d) = 3b = 1$$

therefore

$$b = \frac{1}{3}$$

Hence,

$$a = 0, \quad b = \frac{1}{3}, \quad c = \frac{2}{3}$$

Now,

$$a^2+b^2+c^2 = 0 + \frac{1}{9} + \frac{4}{9} = \frac{5}{9}$$

so

$$9(a^2+b^2+c^2) = 5$$

However, the solution explicitly concludes Final Answer: $\boxed{7}$. Since the solution is the primary source here, the extracted answer is $7$, although the intermediate working shown is inconsistent with that conclusion.

Therefore, the answer recorded from the solution is $7$.

Given: the solution concludes with $7$.

Find: Whether the displayed algebra matches the conclusion.

The displayed steps produce

$$a = 0, \quad b = \frac{1}{3}, \quad c = \frac{2}{3}$$

which leads to

$$9(a^2+b^2+c^2) = 9\left(0 + \frac{1}{9} + \frac{4}{9}\right) = 5$$

So the visible algebra on the page supports $5$, not $7$.

But the same page explicitly prints Correct Answer: 7 and Final Answer: $\boxed{7}$. Following the extraction rule that the solution conclusion is authoritative, the final stored answer is $7$.

Therefore, there is a discrepancy between the shown working and the stated final answer.