Question

The first term of an infinite G.P. is the value of $x$ satisfying the equation $\log_{4}\left(4^{x} - 15\right)+x-2=0$ and the common ratio is $cos\left(\frac{2011 \pi }{3}\right).$ The sum of G.P. is:

• A. $1$
• B. $\frac{4}{3}$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

We have,
$\log_{4}\left(4^{x} - 15\right)+x-2=0$
$\Rightarrow \log_{4}\left(4^{x} - 15\right)=2-x$
$\Rightarrow 4^{x}-15=4^{2 - x}$
$\Rightarrow 4^{x}-15=4^{2}4^{- x}$
$\Rightarrow 4^{2 x}-15\cdot \left(4^{x}\right)=16$
$\Rightarrow 4^{2 x}-15\cdot \left(4^{x}\right)-16=0$
$\Rightarrow 4^{2 x}-16\cdot \left(4^{x}\right)+\left(4^{x}\right)-16=0$
$\Rightarrow \left(4^{x} - 16\right)\left(4^{x} + 1\right)=0$
$\Rightarrow 4^{x}=16$
$\left[\because \left(4^{x} + 1\right) \neq 0 , \forall x \in R\right]$
$\Rightarrow x=2$
Common ratio $=cos\left(\frac{2011 \pi }{3}\right)=cos\left(670 \pi + \frac{\pi }{3}\right)=\frac{1}{2}$
Therefore, sum of infinite GP is
$\frac{x}{1 - \frac{1}{2}}=\frac{2}{\frac{1}{2}}=4$