JEE Main 2026 — 28 January, Evening Shift — Question 7

Let the common difference be $d$. Then $a = 4 - d$, $\alpha = 4 + d$, $b = 4 + 2d$. Given arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$,

so $\frac{1}{a} + \frac{1}{b} = \frac{5}{8}$. Substitute $a = 4-d$, $b = 4+2d$: $\frac{1}{4-d} + \frac{1}{4+2d} = \frac{5}{8}$. Solve: $\frac{4+2d + 4-d}{(4-d)(4+2d)} = \frac{5}{8}$

$\Rightarrow \frac{8+d}{(4-d)(4+2d)} = \frac{5}{8}$. Cross-multiply: $8(8+d) = 5(4-d)(4+2d)$. Expand: $64 + 8d = 5(16 + 8d - 4d - 2d^2) = 5(16 + 4d - 2d^2) = 80 + 20d - 10d^2$. Bring all terms: $64 + 8d - 80 - 20d + 10d^2 = 0$ $\Rightarrow 10d^2 - 12d - 16 = 0$

$\Rightarrow 5d^2 - 6d - 8 = 0$. Solve: $d = \frac{6 \pm \sqrt{36 + 160}}{10} = \frac{6 \pm 14}{10}$. So $d = 2$ or $d = -\frac{4}{5}$.

Since $a = 4-d > 2$, $d < 2$, so $d = 2$ is invalid (gives $a=2$ not >2). Thus $d = -\frac{4}{5}$. Then $\alpha = 4 + d = 4 - \frac{4}{5} = \frac{16}{5}$, $a = 4 - d = 4 + \frac{4}{5} = \frac{24}{5}$, $b = 4 + 2d = 4 - \frac{8}{5} = \frac{12}{5}$. Equation: $\alpha x^2 - a x + 2(\alpha - 2b) = 0$ becomes $\frac{16}{5}x^2 - \frac{24}{5}x + 2\left(\frac{16}{5} - \frac{24}{5}\right) = 0$ $\Rightarrow \frac{16}{5}x^2 - \frac{24}{5}x + 2\left(-\frac{8}{5}\right) = 0$

$\Rightarrow \frac{16}{5}x^2 - \frac{24}{5}x - \frac{16}{5} = 0$. Multiply by 5: $16x^2 - 24x - 16 = 0$ $\Rightarrow 2x^2 - 3x - 2 = 0$. Solve: $x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$.

So $x = 2$ or $x = -\frac{1}{2}$. Roots: $2$ lies in $(1,4)$ and $-\frac{1}{2}$ lies in $(-2,0)$.

Hence option A is correct.