Question

If for $x, y \in R, x>0$,
$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{19}+\ldots . .$ upto $\infty$ terms and $\frac{2+4+6+\ldots+2 y}{3+6+9+\ldots .+3 y}=\frac{4}{\log _{10} x}$, then the ordered pair $( x , y )$ is equal to :

• A. $\left(10^{6}, 6\right)$
• B. $\left(10^{4}, 6\right)$
• C. $\left(10^{2}, 3\right)$
• D. $\left(10^{6}, 9\right)$

Answer 1

Answer: (D)

$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots \infty$
$=\log _{10}\left(x \cdot x^{1 / 3} \cdot x^{1 / 9} \ldots \infty\right)$
$=\log _{10}\left(x^{1+\frac{1}{3}+\frac{1}{9} \ldots \infty}\right)$
$=\log _{10}\left(\frac{1}{1-\frac{1}{3}}\right)$
$\left.=\log _{10} x^{3 / 2}\right)$
$\therefore y=\frac{3}{2} \log _{10} x$
Now,
$2+4+6+\ldots+2 y$
$\frac{2+6+9+\ldots+3 y}{3+6}=\frac{4}{\log _{10} x}$
$\Rightarrow \frac{2(1+2+3+\ldots+y)}{3(1+2+3+\ldots+y)}=\frac{4}{\log _{10} x}$
$\Rightarrow \frac{2}{3}=\frac{4}{\log _{10} x}$
$\Rightarrow \log _{10} x=6$
$\Rightarrow x=10^{6}$
$\therefore y=\frac{3}{2} x 6=9$