- Home
- Standard 11
- Mathematics
7.Binomial Theorem
medium
If the coefficient of the second, third and fourth terms in the expansion of ${(1 + x)^n}$ are in $A.P.$, then $n$ is equal to
A
$7$
B
$2$
C
$6$
D
None of these
(IIT-1994)
Solution
(a) In the expansion of ${(1 + x)^n}$, it is given that $^n{C_1}{,^n}{C_2}{,^n}{C_3}$ are in $A.P.$
==> ${2.^n}{C_2} = {\,^n}{C_1} + {\,^n}{C_3}$
==> $2.\frac{{n(n – 1)}}{{1.2}} = \frac{n}{1} + \frac{{n(n – 1)(n – 2)}}{{1.2.3}}$
==> $6(n – 1) = 6 + (n – 2)(n – 1)$
==> ${n^2} – 9n + 14 = 0$
==> $n = 2$or $n = 7$.
But $n = 2$ is not acceptable because, when $n=2$, there are only three terms in the expansion of ${(1 + x)^2}$, $\therefore $ $n = 7$.
Standard 11
Mathematics
