# If the coefficien

For (1 + x)2n

a=1, b=x and N=2n

We have, For the 2nd term, r=1   ……… Therefore, the coefficient of 2nd term = (2n)

For the 3rd term, r=2    ……….(n! = n. (n-1)!) Therefore, the coefficient of 3rd term = (n)(2n-1)

For the 4th term, r=3    ……….(n! = n. (n-1)!)  Therefore, the coefficient of 3rd term As the coefficients of 2nd, 3rd and 4th terms are in A.P.

Therefore,

2×coefficient of 3rd term = coefficient of 2nd term + coefficient of the 4th term Dividing throughout by (2n),  • 3 (2n-1) = 3 + (2n-1)(n-1)

• 6n – 3 = 3 + (2n2 - 2n – n + 1)

• 6n – 3 = 3 + 2n2 - 3n + 1

• 3 + 2n2 - 3n + 1 - 6n + 3 = 0

• 2n2 - 9n + 7 = 0

Conclusion : If the coefficients of 2nd, 3rd and 4th terms of (1 + x)2n are in A.P. then 2n2 - 9n + 7 = 0

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