Question

If in the expansion of $(1+y{)}^n$, the coefficient of $5^{th},6^{th}$ and $7^{th}$ terms are in A.P., then $n$ is equal to:

• A. $7,11$
• B. $7,14$
• C. $8,16$
• D. none of the above

Answer 1

The correct option is B $7,14$

Given

$(1+y{)}^n={}^n{C}_0+{}^n{C}_{1}y+{}^n{C}_2{y}^2+......+{}^n{C}_n{y}^n$

Here Coefficient of $5^{th},6^{th},7^{th}$ are ${}^n{C}_4,{}^n{C}_5,{}^n{C}_6$ respectively

Since ${}^n{C}_4,{}^n{C}_5,{}^n{C}_6$ are in A.P

${}^n{C}_5=\frac{{}^n{C}_4+{}^n{C}_6}{2}$

$2=\frac{{}^n{C}_4+{}^n{C}_6}{{}^n{C}_5}$

$2=\frac{{}^n{C}_4}{{}^n{C}_5}+\frac{{}^n{C}_6}{{}^n{C}_5}$

We know that

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$

Hence

$2=\frac{5}{n-5+1}+\frac{n-6+1}{6}$

$2=\frac{5\left(6\right)+(n-5)(n-4)}{6(n-4)}$

$12(n-4)=30+n^2-4n-5n+20$

$12n-48=50+n^2-9n$

$n^2-21n+98=0$

$(n-7)(n-14)=0$

$n=7,14$