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7.Binomial Theorem
hard
The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$. Then $n=$
A
$5$
B
$6$
C
$7$
D
$8$
(IIT-2013)
Solution
${ }^{n+5} C_{r-1}:{ }^{n+5} C_r:{ }^{n+5} C_{r+1}=5: 10: 14 $
$\Rightarrow \quad \frac{{ }^{n+5} C_r}{{ }^{n+5} C_{r-1}}=\frac{10}{5} \quad \& \quad \frac{{ }^{n+5} C_{r+1}}{{ }^{n+5} C_r}=\frac{14}{10} $
$\Rightarrow \quad \frac{(n+5)-r+1}{r}=2 \quad \& \quad \frac{(n+5)-(r+1)+1}{r+1}=\frac{7}{5}$
$\Rightarrow \quad \frac{n+6}{r}=3 \quad \& \quad \frac{n+6}{r+1}=\frac{12}{5} $
$\Rightarrow \quad 3 r=\frac{12}{5}(r+1) $
$\Rightarrow \quad r=4 $
$\therefore \quad n+6=12 \quad \Rightarrow \quad n=6 $
Standard 11
Mathematics
