Question

If the $7^{th}$ and $8^{th}$ term of the binomial expansion $(2a - 3b)^n$ are equal, then $\frac{2a + 3b}{2a - 3b}$ is equal to

• A. $\frac{13-n}{n+1}$
• B. $\frac{n+1}{13-n}$
• C. $\frac{6-n}{13-n}$
• D. $\frac{n - 1}{13-n}$
• E. none of the above

Answer 1

Answer: (E)

We have, $(2 a-3 b)^{n}$
$ T_{7} =T_{8} $
$\Rightarrow { }^{n} C_{6}(2 a)^{n-6}(-3 b)^{6} ={ }^{n} C_{7}(2 a)^{n-7}(-3 b)^{7} $
$ \Rightarrow { }^{n} C_{6}(2 a) ={ }^{n} C_{7}(-3 b)$
$ \Rightarrow \frac{2 a}{3 b} =-\frac{{ }^{n} C_{7}}{{ }^{n} C_{6}}$
$\Rightarrow \frac{2 a}{3 b}=-\frac{\frac{n !}{(n-7) ! 7 !}}{\frac{n !}{(n-6) ! 6 !}} $
$\Rightarrow \frac{2 a}{3 b}=-\frac{n-6}{7} $
$ \Rightarrow \frac{2 a}{3 b}=\frac{6-n}{7}$
On applying componendo and dividendo, we get
$\frac{2 a+3 b}{2 a-3 b}=\frac{6-n+7}{6-n-7}$
$=\frac{13-n}{-(n+1)}$
$=-\left[\frac{13-n}{n+1}\right]$