If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$, $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Binomial Theorem

If the coefficients of the middle terms in the binomial expansions of

(1 + α x)^{26}

and

(1 - α x)^{28}

α \neq = 0

, are equal, then the value of

α

is:

Answer: (D)

The middle term of

(1 + α x)^{26}

T_{14} = (13 26) (α x)^{13}

, with coefficient

(13 26) α^{13}

, and the middle term of

(1 - α x)^{28}

T_{15} = (14 28) (- α x)^{14}

, with coefficient

(14 28) α^{14}

. Equating them,

(13 26) = (14 28) α

. Since

(14 28) = \frac{28 \cdot 27}{14 \cdot 14} (13 26) = \frac{27}{7} (13 26)

, this gives

α = \frac{7}{27}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option