Q.8. If the coefficients of the middle terms in the binomial expansions of and , , are equal, then the value of is:

Sol. $(1 + \alpha x)^{26}$ Middle term, $T_{14} = ^{26}C_{13}(\alpha x)^{13}$ For $(1 - \alpha x)^{28}$ Middle term $= ^{28}C_{14}(-\alpha x)^{14} = ^{28}C_{14}(\alpha x)^{14}$ $^{26}C_{13}(\alpha x)^{13} = ^{28}C_{14}(\alpha x)^{14}$ $\Rightarrow ^{26}C_{13} = ^{28}C_{14} \alpha$ $\Rightarrow \alpha = \frac{^{26}C_{13}}{^{28}C_{14}} = \frac{26!}{13!13!} \cdot \frac{14!14!}{28!} = \frac{26! \cdot 14 \cdot 13! \cdot 14 \cdot 13!}{13!13! \cdot 28 \cdot 27 \cdot 26!} = \frac{14 \cdot 14}{28 \cdot 27} = \frac{7}{27}$

What is the correct answer for JEE Main Mathematics Q8?

The correct answer is D. Sol. $(1 + \alpha x)^{26}$ Middle term, $T_{14} = ^{26}C_{13}(\alpha x)^{13}$ For $(1 - \alpha x)^{28}$ Middle term $= ^{28}C_{14}(-\alpha x)^{14} = ^{28}C_{14}(\alpha x)^{14}$ $^{26}C_{13}(\alpha x)^{13} = ^{28}C_{14}(\alpha x)^{14}$ $\Rightarrow ^{26}C_{13} = ^{28}C_{14} \alpha$ $\Rightarrow \alpha = \frac{^{26}C_{13}}{^{28}C_{14}} = \frac{26!}{13!13!} \cdot \frac{14!14!}{28!} = \frac{26! \cdot

Q.8. If the coefficients of the middle terms in the binomial expansions of and , , are equal, then the value of is:

The correct answer is D. Sol. $(1 + \alpha x)^{26}$ Middle term, $T_{14} = ^{26}C_{13}(\alpha x)^{13}$ For $(1 - \alpha x)^{28}$ Middle term $= ^{28}C_{14}(-\alpha x)^{14} = ^{28}C_{14}(\alpha x)^{14}$ $^{26}C_{13}(\alpha x)^{13} = ^{28}C_{14}(\alpha x)^{14}$ $\Rightarrow ^{26}C_{13} = ^{28}C_{14} \alpha$ $\Rightarrow \alpha = \frac{^{26}C_{13}}{^{28}C_{14}} = \frac{26!}{13!13!} \cdot \frac{14!14!}{28!} = \frac{26! \cdot

What are the options for this JEE Main Mathematics question?

A. 1 · B. $\frac{14}{13}$ · C. $\frac{27}{7}$ · D. $\frac{7}{27}$

JEE Main 2026… PYQ · With Answer · Free

JEE Main•Mathematics•MCQs

Q.8. If the coefficients of the middle terms in the binomial expansions of $(1 + ax)^{26}$ and $(1 - ax)^{28}$, $a \ne 0$, are equal, then the value of $a$ is:

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