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If $r$ th and $(r+1)$ th terms in the expansion of $(p+q)^{n}$ are equal, then $\frac{(n+1) q}{r(p+q)}$ is
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The correct answer is:
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Given, $(p+q)^{n}$
$$
T_{r}=T_{(r-1)+1}={ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}
$$
and
$$
T_{r+1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}
$$
From question,
${ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}$
$\frac{n !}{(r-1) !(n-r+1)(n-r) !} \cdot p^{n-r} q^{r} \cdot \frac{p}{q}$
$=\frac{n !}{r(r-1) !(n-r) !} \cdot p^{n-r} \cdot q^{r}$
$\Rightarrow \quad \frac{1}{(n-r+1)} \cdot \frac{p}{q}=\frac{1}{r}$
$\Rightarrow \quad p r=q n-q r+q$
$\Rightarrow \quad \frac{q(n+1)}{r(p+q)}=1$
$$
T_{r}=T_{(r-1)+1}={ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}
$$
and
$$
T_{r+1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}
$$
From question,
${ }^{n} C_{r-1} p^{n-r+1} \cdot q^{r-1}={ }^{n} C_{r} p^{n-r} \cdot q^{r}$
$\frac{n !}{(r-1) !(n-r+1)(n-r) !} \cdot p^{n-r} q^{r} \cdot \frac{p}{q}$
$=\frac{n !}{r(r-1) !(n-r) !} \cdot p^{n-r} \cdot q^{r}$
$\Rightarrow \quad \frac{1}{(n-r+1)} \cdot \frac{p}{q}=\frac{1}{r}$
$\Rightarrow \quad p r=q n-q r+q$
$\Rightarrow \quad \frac{q(n+1)}{r(p+q)}=1$
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