Question

Let n be a positive integer . If the coefficients of 2nd,3rd and 4th terms in the expansion of (1+x)^n are in A.P., find the value of n.

please answer fast.

Answer 1

Basic Binomial Theorem knowledge is required for this question. Ask if you have any questions about the steps.
Hope this helps.

Answer 2

Answer:

Value of n = 7

Step-by-step explanation:

$^nC_{1}, ^nC_{2},^nC_{3} \: are \:\\ coefficients\: of \:2^{nd}, 3^{rd} \: and\: 4^{th} \: terms \\\: in \: the \: expansion \: of (1+x)^{n}\: are\\ \: in \: A.P$

$n, \frac{n(n-1)}{2} \: and \frac{n(n-1)(n-2)}{6} \\\: are \: in \: A.P$

$1,\frac{n-1}{2}\: and \: \frac{n^2-3n+2}{6} \\\: are \: in \: A.P$

/* We know that,

If a,b,c are in A.P then

a+c = 2b */

$1+\frac{n^2-3n+2}{6}=\frac{2(n-1)}{2}$

$\implies \frac{6+n^{2}-3n+2}{6}=n-1$

$\implies n^{2}-3n+8=6(n-1)$

$\implies n^{2}-3n+8=6n-6$

$\implies n^{2}-3n+8-6n+6=0$

$\implies n^{2}-9n+14=0$

/* Splitting the middle term, we get

$\implies n^{2}-7n-2n+14=0$

$\implies n(n-7)-2(n-7)=0$

$\implies (n-7)(n-2)=0$

$\implies (n-7)=0 \: Or \: (n-2)=0$

$\implies n=7\: Or \: n=2$

Here , n=2 is not possible because number of terms are more than three (According to the problem )

Therefore,

n = 7

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