Q70 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
If the third and fourth terms in the expansion $\left(2 x+\dfrac{1}{8}\right)^{10}$ are equal, then the value of $x$ is
- A$\dfrac{1}{8}$
- Correct answerB$\dfrac{1}{6}$
- C$\dfrac{8}{3}$
- D$\dfrac{2}{7}$
Solution
The general term in the expansion of $(a + b)^{n}$ is given by $T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}$.
For the expansion $(2x + \dfrac{1}{8})^{10}$, the third term $T_3$ (where $r=2$) is:
$T_3 = ^{10}C_{2} (2x)^{10-2} (\dfrac{1}{8})^{2} = ^{10}C_{2} (2x)^{8} (\dfrac{1}{8})^{2}$
The fourth term $T_4$ (where $r=3$) is:
$T_4 = ^{10}C_{3} (2x)^{10-3} (\dfrac{1}{8})^{3} = ^{10}C_{3} (2x)^{7} (\dfrac{1}{8})^{3}$
Given $T_3 = T_4$, we have:
$^{10}C_{2} (2x)^{8} (\dfrac{1}{8})^{2} = ^{10}C_{3} (2x)^{7} (\dfrac{1}{8})^{3}$
Dividing both sides by $(2x)^{7} (\dfrac{1}{8})^{2}$:
$^{10}C_{2} (2x) = ^{10}C_{3} (\dfrac{1}{8})$
Calculating the combinations: $^{10}C_{2} = \dfrac{10 \times 9}{2} = 45$ and $^{10}C_{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
$45(2x) = 120(\dfrac{1}{8})$
$90x = 15$
$x = \dfrac{15}{90} = \dfrac{1}{6}$
Answer: $\dfrac{1}{6}$
For the expansion $(2x + \dfrac{1}{8})^{10}$, the third term $T_3$ (where $r=2$) is:
$T_3 = ^{10}C_{2} (2x)^{10-2} (\dfrac{1}{8})^{2} = ^{10}C_{2} (2x)^{8} (\dfrac{1}{8})^{2}$
The fourth term $T_4$ (where $r=3$) is:
$T_4 = ^{10}C_{3} (2x)^{10-3} (\dfrac{1}{8})^{3} = ^{10}C_{3} (2x)^{7} (\dfrac{1}{8})^{3}$
Given $T_3 = T_4$, we have:
$^{10}C_{2} (2x)^{8} (\dfrac{1}{8})^{2} = ^{10}C_{3} (2x)^{7} (\dfrac{1}{8})^{3}$
Dividing both sides by $(2x)^{7} (\dfrac{1}{8})^{2}$:
$^{10}C_{2} (2x) = ^{10}C_{3} (\dfrac{1}{8})$
Calculating the combinations: $^{10}C_{2} = \dfrac{10 \times 9}{2} = 45$ and $^{10}C_{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
$45(2x) = 120(\dfrac{1}{8})$
$90x = 15$
$x = \dfrac{15}{90} = \dfrac{1}{6}$
Answer: $\dfrac{1}{6}$
Hence, the correct answer is option: 'B'.