Question

If the coefficients of 6th and 5th terms of expansion $(1+x{)}^n$ are in the ratio 7:5, then find the value of n.

• A. $11$
• B. $12$
• C. $10$
• D. $9$

Answer 1

The correct option is A $11$

Given,

$(1+x{)}^n$

General term

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^n{C}_r(1{)}^{n-r}{x}^r$

Given

$\frac{coeffof{T}_6}{coeffof{T}_5}=\frac{7}{5}$

$T_6=T_{5+1}={}^n{C}_5(1{)}^{n-5}{x}^5$

$T_5=T_{4+1}={}^n{C}_4(1{)}^{n-4}{x}^4$

$\frac{{}^n{C}_5}{{}^n{C}_4}=\frac{7}{5}$

$\frac{n!4!\times (n-4)!}{5!\times (n-5)!\times n!}=\frac{7}{5}$

$\frac{n!\times 4!\times (n-4)\times (n-5)!}{5!\times (n-5)!\times n!}=\frac{7}{5}$

$\frac{(n-4)}{5}=\frac{7}{5}$

$n=7+4$

$n=11$

Hence, correct option is A.