Question

The 3rd, 4th, 5th terms in the expansion of ${(x+a)}^n$ are respectively $84,280$ and $560$. Find the values of $x,a$ and $n$.

Answer 1

Given that

$T_3{=}^n{C}_2{x}^{n-2}{a}^2=84$

$\Rightarrow \frac{n(n-1)}{2}{x}^{n-2}{a}^2=84$ ...$\left(i\right)$

$T_4{=}^n{C}_3{x}^{n-3}{a}^3=280$

$\Rightarrow \frac{n(n-1)(n-2)}{6}{x}^{n-3}{a}^3=280$ ...$\left(ii\right)$

$T_5{=}^n{C}_4{x}^{n-4}{a}^4=560$

$\Rightarrow \frac{n(n-1)(n-2)(n-3)}{24}{x}^{n-4}{a}^4=560$ ...$\left(iii\right)$

$\Rightarrow \frac{T_3\times T_4}{T_4^2}=\frac{84\times 560}{{280}^2}=\frac{3(n-3)}{4(n-2)}$

$\Rightarrow \frac{3}{5}=\frac{3(n-3)}{4(n-2)}$

$\Rightarrow 5n-15=4n-18$

$\Rightarrow n=7$ ...$\left(iv\right)$

Divide $\left(ii\right)$ by $\left(i\right)$ we get

$\Rightarrow \frac{n-2}{3}.\frac{a}{x}=\frac{280}{84}$

$\Rightarrow \frac{7-2}{3}\times \frac{a}{x}=\frac{280}{84}$

$\Rightarrow x=\frac{a}{2}$ ...$\left(v\right)$

Using $\left(iv\right)$ and $\left(v\right)$

$\Rightarrow 84=\frac{7(7-1)}{2}(\frac{a}{2}{)}^{7-2}{a}^2$

$\Rightarrow 84=\frac{a^7}{2^5}.\left(21\right)$

$\Rightarrow a^7=2^7$

$\Rightarrow a=2$

$\Rightarrow x=\frac{a}{2}=\frac{2}{2}=1$

Thus $x=1;a=2;n=7$