Question

If 3${}^{rd}$ ,4${}^{th}$ and 5${}^{th}$ terms in the expansion of $(x+A{)}^n$ are respectively $84,280,560$, then the set of value of $x$, $A$ and $n$ are respectively

• A. $(1,2,7)$
• B. $(1,2,5)$
• C. $(2,1,5)$
• D. $(7,2,5)$

Answer 1

The correct option is A $(1,2,7)$

As per the given conditions.
$T_3={}^n{C}_2{x}^{n-2}{a}^2=84$
$T_4={}^n{C}_3{x}^{n-3}{a}^3=280$
$T_5={}^n{C}_4{x}^{n-4}{a}^4=560$
Hence
$3{}^n{C}_3{x}^{n-3}{a}^3=10{}^n{C}_2{x}^{n-2}{a}^2$
$3\left(a\right)(n-2)=10\left(3\right)\left(x\right)$
$\frac{x}{a}=\frac{n-2}{10}$ ...(i)
$T_5=2T_4$
$\frac{n!}{(n-4)!.4!}a=2\frac{n!}{(n-3)!.4!}.x$
$\frac{n-3}{8}=\frac{x}{a}$ ...(ii)
Comparing i and ii we get
$5(n-3)=4(n-2)$
$5n-15=4n-8$
$n=7$
Hence answer is A.