Question

If the coefficient of the $5th,6th$ and $7th$ terms of the expansion of $(1+x{)}^n$ are in A.P then the value of $n$may be?

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C

$7$

Explanation for the correct option.

Step 1: Relation between coefficients

Given coefficient of the $5th,6th$ and $7th$ terms of the expansion of $(1+x{)}^n$ are in A.P.

Using the condition of A.P we have:

$\begin{array}{rcl}{2}^{\mathrm{n}}{\mathrm{C}}_5& =& {}^{\mathrm{n}}\mathrm{C}_4+{}^{\mathrm{n}}\mathrm{C}_6\\ 2\frac{\mathrm{n}!}{5!(\mathrm{n}-5)!}& =& \frac{\mathrm{n}!}{4!(\mathrm{n}-4)!}+\frac{\mathrm{n}!}{6!(\mathrm{n}-6)!}\\ \frac{2}{5(\mathrm{n}-5)}& =& \frac{1}{(\mathrm{n}-4)(\mathrm{n}-5)}+\frac{1}{30}\\ \frac{2}{5(\mathrm{n}-5)}-\frac{1}{(\mathrm{n}-4)(\mathrm{n}-5)}& =& \frac{1}{30}\\ \frac{2(\mathrm{n}-4)-5}{5(\mathrm{n}-4)(\mathrm{n}-5)}& =& \frac{1}{30}\\ \frac{2\mathrm{n}-13}{{\mathrm{n}}^2-9\mathrm{n}+20}& =& \frac{1}{6}\end{array}$

Step 2: Find the value of $n$

Using cross multiplication we have:

$\begin{array}{rcl}{\mathrm{n}}^2-9\mathrm{n}+20& =& 12\mathrm{n}-78\\ {\mathrm{n}}^2-21\mathrm{n}+98& =& 0\\ {\mathrm{n}}^2-14\mathrm{n}-7\mathrm{n}+98& =& 0\\ \mathrm{n}(\mathrm{n}-14)-7(\mathrm{n}-14)& =& 0\\ (\mathrm{n}-14)(\mathrm{n}-7)& =& 0\\ \mathrm{n}& =& 14\text{or}\mathrm{n}=7\end{array}$

Hence, option C is correct.