Question

the coefficient of the 5th, 6th and 7th terms in a binomial expansion of (1+x)^n in ascending powers of x are consecutive terms in a linear sequence. Find the possible values of n​

Answer 1

Answer:

Step-by-step explanation:

given :

• the coefficient of the 5th, 6th and 7th terms in a binomial expansion of (1+x)^n in ascending powers of x are consecutive terms in a linear sequence. Find the possible values of n

to find :

• Find the possible values of n

solution :

• hence,the answer is 7

Answer 2

$\large\underline{\sf{Solution-}}$

Given expansion is

$\red{\rm :\longmapsto\: {(1 + x)}^{n}}$

We know,

$\sf \:The \: coefficient \: of \: {(r + 1)}^{th} \: term \: of \: {(1 + x)}^{n} = \: ^nC_r$

So,

$\red{\rm :\longmapsto\:Coefficient \: of \: {5}^{th} \: term \: = \: ^nC_4}$

$\red{\rm :\longmapsto\:Coefficient \: of \: {6}^{th} \: term \: = \: ^nC_5}$

$\red{\rm :\longmapsto\:Coefficient \: of \: {7}^{th} \: term \: = \: ^nC_6}$

Since, the coefficient of the 5th, 6th and 7th terms in a binomial expansion of (1+x)^n in ascending powers of x are consecutive terms in a linear sequence.

Its mean coefficient of 5th, 6th and 7th terms form an AP series.

We know,

• If a, b, c are in AP, then 2b = a + c.

So,

$\rm \implies\:2^nC_5 \: = \: ^nC_4 \: + \: ^nC_6$

$\rm :\longmapsto\:2\dfrac{n!}{5! \: (n - 5)!} = \dfrac{n!}{4! \: (n - 4)!} + \dfrac{n!}{6! \: (n - 6)!}$

$\rm :\longmapsto\:\dfrac{2}{5.4! \: (n - 5)(n - 6)!} = \dfrac{1}{4! \: (n - 4)(n - 5)(n - 6)!} + \dfrac{n!}{6.5.4! \: (n - 6)!}$

$\rm :\longmapsto\:\dfrac{2}{5(n - 5)} = \dfrac{1}{(n - 4)(n - 5)} + \dfrac{1}{30}$

$\rm :\longmapsto\:\dfrac{2}{5(n - 5)} - \dfrac{1}{(n - 4)(n - 5)} = \dfrac{1}{30}$

$\rm :\longmapsto\: \dfrac{2( n - 4) - 5}{5(n - 4)(n - 5)} = \dfrac{1}{30}$

$\rm :\longmapsto\: \dfrac{2n - 8- 5}{(n - 4)(n - 5)} = \dfrac{1}{6}$

$\rm :\longmapsto\: \dfrac{2n - 13}{ {n}^{2} - 9n + 20} = \dfrac{1}{6}$

$\rm :\longmapsto\: {n}^{2} - 9n + 20 = 12n - 78$

$\rm :\longmapsto\: {n}^{2} - 9n + 20 - 12n + 78 = 0$

$\rm :\longmapsto\: {n}^{2} - 21n + 98 = 0$

$\rm :\longmapsto\: {n}^{2} - 14n - 7n + 98 = 0$

$\rm :\longmapsto\:n(n - 14) - 7(n - 14) = 0$

$\rm :\longmapsto\:(n - 14)(n - 7) = 0$

$\bf\implies \:n = 14 \: \: or \: \: n = 7$