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The two successive temrs in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio $1: 4$ are
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$5^{\text {th }}$ and $6^{\text {th }}$
$5^{\text {th }}$ and $6^{\text {th }}$
Suppose two successive terms in the exspansion of $(1+x)^{24}$ are $(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ terms.
So, $T_{r+1}={ }^{24} C_r x^r$ and $T_{r+2}={ }^{24} C_{r+1} x^{r+1}$
$$
\begin{aligned}
&\text { As, } \frac{{ }^{24} C_r}{{ }^{24} C_{r+1}}=\frac{1}{4} \Rightarrow \frac{\frac{(24) !}{r !(24-r) !}}{\frac{(24) !}{(r+1) !(24-r-1) !}}=\frac{1}{4} \\
&\Rightarrow \frac{r+1}{24-r}=\frac{1}{4} \Rightarrow 5 r=20 \Rightarrow r=4
\end{aligned}
$$
Therefore, required terms are $5^{\text {th }}$ and $6^{\text {th }}$ terms.
So, $T_{r+1}={ }^{24} C_r x^r$ and $T_{r+2}={ }^{24} C_{r+1} x^{r+1}$
$$
\begin{aligned}
&\text { As, } \frac{{ }^{24} C_r}{{ }^{24} C_{r+1}}=\frac{1}{4} \Rightarrow \frac{\frac{(24) !}{r !(24-r) !}}{\frac{(24) !}{(r+1) !(24-r-1) !}}=\frac{1}{4} \\
&\Rightarrow \frac{r+1}{24-r}=\frac{1}{4} \Rightarrow 5 r=20 \Rightarrow r=4
\end{aligned}
$$
Therefore, required terms are $5^{\text {th }}$ and $6^{\text {th }}$ terms.
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