Question

If the ratio of the coefficient of third and fourth term in the expansion of $(x-\frac{1}{2x}{)}^n$ is 1:2, then the value of n will be

• A. 18
• B. 16
• C. 12
• D. -10

Answer 1

The correct option is D

-10

$T_3$ = ${}^n{C}_2$$(x{)}^{n-2}$$(-\frac{1}{2x}{)}^2$ and $T_4$ = ${}^n{C}_3$$(x{)}^{n-3}$$(-\frac{1}{2x}{)}^3$

But according to the condition,

$\frac{-n(n-1)\times 3\times 2\times 1\times 8}{n(n-1)(n-2)\times 2\times 1\times 4}$ = $\frac{1}{2}$ ⇒ n = -10