Binomial TheoremmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Equal Middle-Term Coefficients: α = 7/27 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

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

(1+\alpha x)^{26}

and

(1-\alpha x)^{28}

\alpha\ne0

, are equal, then the value of

\alpha

1

\dfrac{14}{13}

\dfrac{27}{7}

\dfrac{7}{27}

correct

Solution

Step 1:

(1+\alpha x)^{26}

T_{r+1}={}^{26}C_{r}(\alpha x)^{r}

; middle term is the

14

-th,

r=13

T_{14}={}^{26}C_{13}\,\alpha^{13}x^{13}.

(1-\alpha x)^{28}

T_{r+1}={}^{28}C_{r}(-\alpha x)^{r}

; middle term is the

15

-th,

r=14

T_{15}={}^{28}C_{14}(-1)^{14}\alpha^{14}x^{14}={}^{28}C_{14}\,\alpha^{14}x^{14}.

Step 2:

{}^{26}C_{13}\,\alpha^{13}={}^{28}C_{14}\,\alpha^{14}

. Divide by

\alpha^{13}

(

\alpha\ne0

\alpha=\dfrac{{}^{26}C_{13}}{{}^{28}C_{14}}.

Step 3:

\alpha=\dfrac{26!}{13!\,13!}\cdot\dfrac{14!\,14!}{28!}=\dfrac{26!}{28!}\cdot\dfrac{14!}{13!}\cdot\dfrac{14!}{13!}=\dfrac{1}{28\cdot27}\cdot14\cdot14=\dfrac{196}{756}=\dfrac{7}{27}.

Correct answer: (4)

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