Question:
If for $x, y \in \mathbf{R}, x>0$
$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots .$ 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 $(\mathrm{x}, \mathrm{y})$ is equal to :
Correct Option: , 4
Solution:
$\frac{2(1+2+3+\ldots+y)}{3(1+2+3+\ldots+y)}=\frac{4}{\log _{10} x}$
$\Rightarrow \log _{10} x=6 \Rightarrow x=10^{6}$
Now,
$y=\left(\log _{10} x\right)+\left(\log _{10} x^{\frac{1}{3}}\right)+\left(\log _{10} x^{\frac{1}{9}}\right)+. . \infty$
$=\left(1+\frac{1}{3}+\frac{1}{9}+\ldots \infty\right) \log _{10} x$
$=\left(\frac{1}{1-\frac{1}{3}}\right) \log _{10} x=9$
So, $(x, y)=\left(10^{6}, 9\right)$