Question

Given that the $4^{th}$ term in the expansion of $(2+\frac{3}{8}a{)}^{10}$ has the maximum numerical value, then

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Consider the expansion of $(x+y{)}^n$. Let us assume $T_{r+1}$ has the numerically greatest term. Then, $|T_{r+1}|\ge |T_r|$

$\Rightarrow |\frac{T_{r+1}}{T_r}|\ge 1$
$\Rightarrow |\frac{{}^n{C}_r{x}^{n-r}{y}^r}{{}^n{C}_{r-1}{x}^{n-r+1}{y}^{r-1}}|\ge 1$

$\Rightarrow \frac{n-r+1}{r}|\frac{y}{x}|\ge 1$

In our question, x=2, y = $\frac{3a}{8}$ and n=10
$\Rightarrow \frac{10-r+1}{r}|\frac{3a}{8\times 2}|\ge 1$

It is given that $T_4$ is the numerically greatest term
$\Rightarrow$ r=3($T_{r+1}$ is the numerically greatest term, not $T_r$)
$\Rightarrow \frac{10-3+1}{3}|\frac{3a}{8\times 2}|\ge 1$

$\Rightarrow |\frac{3a}{8\times 2}|\ge \frac{3}{8}$

$\Rightarrow |a|\ge 2$