Question

In the expansion of ${(\sqrt{y}+\frac{1}{2\sqrt[4]{4y}})}^n$, the first three co-efficients form an A.P. then $n=$

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

8

The first coefficient in the above expansion would be $1$
The second coefficient in the above expansion will be $\frac{{}^n{C}_1}{2}$
The third coefficient in the above expansion will be $\frac{{}^n{C}_2}{2^2}$
Since, it is given that they are in A.P.
Therefore, by applying the above condition, we get.
$2\frac{{}^n{C}_1}{2}=1+\frac{{}^n{C}_2}{2^2}$
${}^n{C}_1=1+\frac{{}^n{C}_2}{4}$
$n=1+\frac{n(n-1)}{2!.4}$
$n-1=\frac{n(n-1)}{8}$
$n=8$
Hence, the value of $n$ is $8$.