Question

The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are
(a) 3^rd and 4^th
(b) 4^th and 5^th
(c) 5^th and 6^th
(d) 6^th and 7^th

Answer 1

For (1 + x)²⁴ two successive terms have coefficients in ration 1 : 4
Let the two successive terms be (r + 1)^th and (r + 2) terms
$$\begin{gathered} \mathrm{i}.\mathrm{e}.T_{r+1}{=}^{24}{C}_r{x}^r\mathrm{and}{T}_{r+2}{=}^{24}{C}_{r+1}{x}^{r+1} \\ \mathrm{Given}\mathrm{that}\frac{{}^{24}{C}_r}{{}^{24}{C}_{r+1}}=\frac{1}{4} \\ \mathrm{i}.\mathrm{e}.\frac{\left(24\right)!}{r!(24-r)!}\times \frac{(r+1)!(24-r-1)!}{\left(24\right)!}=\frac{1}{4} \\ \mathrm{i}.\mathrm{e}.\frac{(r+1)r!(23-r)!}{r!(24-r)(23-r)!}=\frac{1}{4} \\ \mathrm{i}.\mathrm{e}.\frac{r+1}{24-r}=\frac{1}{4} \\ \mathrm{i}.\mathrm{e}.4r+4=24-r \\ \mathrm{i}.\mathrm{e}.5r=20 \\ \mathrm{i}.\mathrm{e}.r=4 \\ \mathrm{i}.\mathrm{e}.T_{r+1}=T_{4+1}=T_5\mathrm{and}{T}_{r+2}=T_{4+2}=T_6\mathrm{are}\mathrm{the}\mathrm{two}\mathrm{terms} \end{gathered}$$
Hence 5^th and 6^th terms.
Hence, the correct answer is option C.