In the binomial expansion of x+12x4n,n∈N the coefficients of first, second and third terms form an A.P. Then n =

x+12x4n=∑r=0nCr nxn-r.12x4r =∑r=0nCr n.12rx2n-3r4 given that 1, n2,n(n-1)8 are in A.P ⇒n=1+n(n-1)8 ⇒n=8

In the binomial expansion of x+12x4n,n∈N the coefficients of first, second and third terms form an A.P. Then n =

x+12x4n=∑r=0nCr nxn-r.12x4r =∑r=0nCr n.12rx2n-3r4 given that 1, n2,n(n-1)8 are in A.P ⇒n=1+n(n-1)8 ⇒n=8

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In the binomial expansion of ${(\sqrt{x} + \frac{1}{2 \sqrt[4]{x}})}^{n}, n \in N$ the coefficients of first, second and third terms form an A.P. Then n =

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Detailed Solution

${(\sqrt{x} + \frac{1}{2 \sqrt[4]{x}})}^{n} = \sum_{r = 0}^{n} {}^{n}C_{r} {(\sqrt{x})}^{n - r} . {(\frac{1}{2 \sqrt[4]{x}})}^{r} = \sum_{r = 0}^{n} {}^{n}C_{r} . \frac{1}{2^{r}} x^{\frac{2 n - 3 r}{4}} g i v e n t h a t 1, \frac{n}{2}, \frac{n (n - 1)}{8} are in A . P \Rightarrow n = 1 + \frac{n (n - 1)}{8} \Rightarrow n = 8$