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
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