JEE Advanced · Mathematics · 6. Binomial Theorem
The coefficients of three consecutive terms of are in the ratio . Then
- A 21
- B 34
- C 6
- D 8
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\({ }^{ n +5} C _{ r -1}:{ }^{ n +5} C _{ r }:{ }^{ n +5} C _{ r +1}=5: 10: 14 \)
\( \Rightarrow \frac{(n+5)!}{(r-1)!(n-r+6)!}: \frac{(n+5)!}{(r)!(n-r+5)!}:\) \(\frac{(n+5)!}{(r+1)!(n-r+4)!}=5: 10: 14 \)
\( \Rightarrow \frac{1}{(r-1)!(n-r+6)!}: \frac{1}{(r)!(n-r+5)!}:\) \(\frac{1}{(r+1)!(n-r+4)!}=5: 10: 14 \)
\( \Rightarrow \frac{(r+1) r}{(n-r+6)!}: \frac{r+1}{(n-r+5)!}: \frac{1}{(n-r+4)!}=\) \(5: 10: 14 \)
\( \Rightarrow(r+1) r:(r+1)(n-r+6):(n~-\) \(r+6)(n-r+5)=5: 10: 14 \)
\( \Rightarrow \frac{(r+1)(n-r+6)}{(r+1) r}=\frac{10}{5}\) and \(\frac{(n-r+6)(n-r+5)}{(r+1)(n-r+6)}=\frac{14}{10} \)
\( \Rightarrow \frac{(n-r+6)}{r}=2\) and \(\frac{(n-r+5)}{(r+1)}=\frac{7}{5} \)
\( \Rightarrow n=3 r-6\) and \(n =\frac{12 r-18}{5} \)
\( \Rightarrow 3 r-6=\frac{12 r-18}{5}=n \)
\( \Rightarrow 15 r-30=12 r-18\)
\(\Rightarrow r=4\)
\(\Rightarrow n=6\)
\( \Rightarrow \frac{(n+5)!}{(r-1)!(n-r+6)!}: \frac{(n+5)!}{(r)!(n-r+5)!}:\) \(\frac{(n+5)!}{(r+1)!(n-r+4)!}=5: 10: 14 \)
\( \Rightarrow \frac{1}{(r-1)!(n-r+6)!}: \frac{1}{(r)!(n-r+5)!}:\) \(\frac{1}{(r+1)!(n-r+4)!}=5: 10: 14 \)
\( \Rightarrow \frac{(r+1) r}{(n-r+6)!}: \frac{r+1}{(n-r+5)!}: \frac{1}{(n-r+4)!}=\) \(5: 10: 14 \)
\( \Rightarrow(r+1) r:(r+1)(n-r+6):(n~-\) \(r+6)(n-r+5)=5: 10: 14 \)
\( \Rightarrow \frac{(r+1)(n-r+6)}{(r+1) r}=\frac{10}{5}\) and \(\frac{(n-r+6)(n-r+5)}{(r+1)(n-r+6)}=\frac{14}{10} \)
\( \Rightarrow \frac{(n-r+6)}{r}=2\) and \(\frac{(n-r+5)}{(r+1)}=\frac{7}{5} \)
\( \Rightarrow n=3 r-6\) and \(n =\frac{12 r-18}{5} \)
\( \Rightarrow 3 r-6=\frac{12 r-18}{5}=n \)
\( \Rightarrow 15 r-30=12 r-18\)
\(\Rightarrow r=4\)
\(\Rightarrow n=6\)
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