If the coefficients of \(r\) th and \((r+1)\) th terms in the expansion of \((1+x)^{24}\) are in the ratio…
If the coefficients of \(r\) th and \((r+1)\) th terms in the expansion of \((1+x)^{24}\) are in the ratio \(12: 13\), then \(r\) is the root of the quadratic equation
\(x^2-5 x+6=0\)
\(x^2-11 x+30=0\)
\(x^2-14 x+13=0\)
\(x^2-14 x+24=0\)
Solution
According to given information,
\(\frac{{ }^{24} C_{r-1}}{{ }^{24} C_r}=\frac{12}{13}\)
\(\begin{gathered}
\Rightarrow \frac{\frac{24 !}{(r-1) !(25-r) !}}{\frac{24 !}{r !(24-r) !}}=\frac{12}{13} \\
\Rightarrow \frac{r}{25-r}=\frac{12}{13} \Rightarrow r=12
\end{gathered}\)
and since, \(x^2-14 x+24=0 \Rightarrow(x-12)(x-2)=0\)
\(\Rightarrow \quad x=2,12\)
Hence, option (d) is correct.