Question

If ${}^n{C}_4$, ${}^n{C}_5$, ${}^n{C}_6$ of $(1+x{)}^n$ are in A.P. then $n=$

• A. 12
• B. 11
• C. 7
• D. 8

Answer 1

The correct option is C 7

Let ${}^n{C}_{r-1},{}^n{C}_r,{}^n{C}_{r+1}$ be in AP.
Therefore condition for the following terms to be in A.P is
$(n-2r{)}^2=n+2$ ...(i)
In the above question
${}^n{C}_{r-1}={}^n{C}_4$
Hence $r-1=4$
$\therefore r=5$
Substituting in (i), we get
$(n-10{)}^2=n+2$
$n^2-20n+100=n+2$
$n^2-21n+98=0$
$(n-14)(n-7)=0$
$n=7$ and $n=14$
Hence, answer is option C