AP in Binomial

The expansion of ( x + 1 2 x 4 ) n \left( \sqrt{x} + \dfrac{1}{2\sqrt[4]{x}}\right)^{n} is arranged in decreasing powers of x x . If coefficient of first three terms form an arithmetic progression , then one of the integral powers of x x in its expansion is can be equal to __________ \text{\_\_\_\_\_\_\_\_\_\_} .

2 8 0 4

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1 solution

Rahil Sehgal
Feb 22, 2018

first notice that the coeff. of x will be highest if the power of second term 1 2 x 4 \dfrac{1}{2\sqrt[4]{x}} will be least so first 3 terms. coeff will be = nC0, 1/2 (nC1) , 1/4 (nC4), solve and get n=8

Hey @Rahil Sehgal !

Ankit Kumar Jain - 3 years, 3 months ago

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hi @Ankit Kumar Jain

Rahil Sehgal - 3 years, 3 months ago

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