Question

The second, third and fourth terms in the binomial expansion $(x+a)^n$ are $240$, $720$ and $1080$, respectively. Find $x$, $a$ and $n$ respectively.

• A. $2$, $3$, $5$
• B. $1$, $3$, $4$
• C. $2$, $3$, $6$
• D. $3$, $4$, $5$

Answer 1

Answer: (A)

$2$, $3$, $5$

Answer 2

We know that $T_{r+1} = \,^{n}C_{r}x^{n-r}\cdot a^{r}$ So, $T_{2} = \,^{n}C_{1}x^{n-1}\cdot a = 240\quad\ldots\left(1\right)$ $T_{3} = \,^{n}C_{2}x^{n-2}\, a^{2} = 720\quad \ldots \left(2\right)$ $T_{4} = \,^{n}C_{3}x^{n-3} \,a^{3} = 1080\quad \ldots \left(3\right)$ Dividing $\left(2\right)$ by $\left(1\right)$, we get $\frac{^{n}C_{2}x^{n-2}a^{2}}{^{n}C_{1}x^{n-1} a} = \frac{720}{240}$ i.e., $\left(n-1\right)\cdot\frac{a}{x} = 6$ or $\frac{a}{x} = \frac{6}{\left(n-1\right)}\quad\ldots\left(4\right)$ Dividing $\left(3\right)$ by $\left(2\right)$, we have $\frac{a}{x} = \frac{9}{2\left(n-2\right)}\quad \ldots \left(5\right)$ From $\left(4\right)$ and $\left(5\right)$, $\frac{6}{n-1} = \frac{9}{2\left(n-2\right)}$. $\therefore n= 5$ Hence, from $\left(1\right)$, $5x^{4}\,a = 240$, and from $\left(4\right)$, $\frac{a}{x} = \frac{3}{2}$ Solving these equations, we get $x = 2$ and $a = 3$.