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  • If the coefficients of 2nd 3rd and the 4th terms in the expansion of (1+x)^n are in A.P., then the value of n is (a) 2 (b) 7 (c) 11 (d) 14

If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then the value of n is
(a) 2           
(b) 7 
(c) 11               
(d) 14

 

Answers (1)

The answer is the option (b) 7

(1+x)n = nC0 + nC1x + nC2x2 + nC3x3 + ….. + nCnxn

Now, nC1, nC2, nC3 are the coefficients of the second, third and fourth terms respectively.

2 nC2 =nC1 + nC3      …… since they are in A.P.

2[n!/(n-2)!2!] = n + n!/ 3!(n-3)!

2[n(n-1)/2!] = n + n(n-1)(n-2)/3!

(n-1) = 1 + (n-1)(n-2)/6

6n – 6 = 6 + n2 – 3n + 2

n2 – 9n + 14 = 0

(n-7)(n-2) = 0

Thus, n = 2 or n = 7

Now, n = 2 is not possible,

Thus, n = 7
 

Posted by

infoexpert22

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