The two successive terms in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio 1: 4 are
(A) $3^{\text {rd }}$ and $4^{\text {th }}$
(B) $4^{\text {th }}$ and $5^{\text {th }}$
(C) $5^{\text {th }}$ and $6^{\text {th }}$
(D) $6^{\text {th }}$ and $7^{\text {th }}$
(C) $5^{\text {th }}$ and $6^{\text {th }}$
Explanation:
Let the two successive terms in the expansion of $(1+x)^{24}$ be $(r+1)^{\text {th }}$ and $(r+$ 2) $^{\text {th }}$ term
Now,
$T_{r+1}={ }^{24} C_{r} x^{r}$ and $T_{r+2}={ }^{24} C_{r+1} x^{r+1}$
Given
$\frac{{ }^{24} C_{r}}{{ }^{24} C_{r+1}}=\frac{1}{4}$
$\Rightarrow \frac{\frac{(24) !}{r !(24-r) !}}{\frac{(24) !}{(r+1) !(24-r-1) !}}=\frac{1}{4} \Rightarrow \frac{(r+1) r !(23-r) !}{r !(24-r)(23-r) !}=\frac{1}{4} \Rightarrow \frac{r+1}{24-r}=\frac{1}{4}$
$4 r+4=24-r$
Which implies $r=4$
$T_{4+1}=T_{5}$ and $T_{4+2}=T_{6}$
Hence the correct option is $\mathrm{c}$