KCET · Maths · Probability
If \( 21^{\text {st }} \) and \( 22^{\text {nd }} \) terms in the expansion of \( (1+x)^{44} \) are equal, then \( x \) is equal to
- A \( \frac{21}{22} \)
- B \( \frac{23}{24} \)
- C \( \frac{8}{7} \)
- D \( \frac{7}{8} \)
Answer & Solution
Correct Answer
(D) \( \frac{7}{8} \)
Step-by-step Solution
Detailed explanation
Given that \( (1+x)^{44} \)
\[
\begin{array}{l}
\text { Now, }{ }^{44} C_{20} x^{20}={ }^{44} C_{21} x^{21} \\
\Rightarrow x=\frac{{ }^{44} C_{20}}{{ }^{44} C_{2}} \\
=\frac{(44-21) ! 21 !}{(44-20) ! 20 !}=\frac{23 !}{24 !} \cdot \frac{21 !}{20 !} \\
=\frac{21}{24}=\frac{7}{8}
\end{array}
\]
\[
\begin{array}{l}
\text { Now, }{ }^{44} C_{20} x^{20}={ }^{44} C_{21} x^{21} \\
\Rightarrow x=\frac{{ }^{44} C_{20}}{{ }^{44} C_{2}} \\
=\frac{(44-21) ! 21 !}{(44-20) ! 20 !}=\frac{23 !}{24 !} \cdot \frac{21 !}{20 !} \\
=\frac{21}{24}=\frac{7}{8}
\end{array}
\]
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