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Question:
If $21^{st}$ and $22^{nd}$ terms in the expansion of $(1 + x)^{44}$ are equal, then $x$ is equal to
If $21^{st}$ and $22^{nd}$ terms in the expansion of $(1 + x)^{44}$ are equal, then $x$ is equal to
Updated On: Oct 9, 2024
- $\frac {21}{22}$
- $\frac {23}{24}$
- $\frac {8}{7}$
- $\frac {7}{8}$
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The Correct Option is D
Solution and Explanation
Given expansion is $(1+x)^{44}$.
According to the question, $T_{21}=T_{22}$
${ }^{44} C_{20} x^{20}={ }^{44} C_{21} x^{21}$
$\Rightarrow x=\frac{{ }^{44} C_{20}}{{ }^{44} C_{21}}$
$\Rightarrow x=\frac{\frac{44 !}{20 ! \times 24 !}}{\frac{44 !}{21 ! \times 23 !}}=\frac{21 ! \times 23 !}{20 ! \times 24 !}$
$=\frac{21}{24}=\frac{7}{8}$
According to the question, $T_{21}=T_{22}$
${ }^{44} C_{20} x^{20}={ }^{44} C_{21} x^{21}$
$\Rightarrow x=\frac{{ }^{44} C_{20}}{{ }^{44} C_{21}}$
$\Rightarrow x=\frac{\frac{44 !}{20 ! \times 24 !}}{\frac{44 !}{21 ! \times 23 !}}=\frac{21 ! \times 23 !}{20 ! \times 24 !}$
$=\frac{21}{24}=\frac{7}{8}$
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