Question

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

• A. $5$
• B. $6$
• C. $7$
• D. $7$ or $14$

Answer 1

The correct option is D $7$ or $14$

$(1+x{)}^n{=}^n{C}_{0}1{+}^n{C}_{1}1x+...{+}^n{C}_n{x}^n$

$5^{th}$ term ${\Rightarrow }^n{C}_4{x}^4$

$6^{th}$ term ${\Rightarrow }^n{C}_5{x}^5$

$7^{th}$ term ${\Rightarrow }^n{C}_6{x}^6$

They are in A.P (coefficient)

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

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

$2\frac{1}{5\times 4!(n-5)(n-6)!}=\frac{1}{4!(n-4)(n-5)(n-6)!}+\frac{1}{6\times 5\times 4!\times (n-6)!}$

$\frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$

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

$n^2-9n-12n+48+50=0$

$n-14n-7n+98=0$

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

$n=14$ or $n=7$

$nϵ7$ or 14 (option D)