Mathematics - Sequence & Series Question with Solution | TestHub

$\frac{2+4+6+\dots +2y}{3+6+9+\dots +3y}=\frac{4}{{\log }_{10}x}$

$\frac{2\left(1+2+3+----y\right)}{3\left(1+2+3+----y\right)}=\frac{4}{{\log }_{10}x}$

${\log }_{10}x=6$

$x={10}^6$

Now

$y={\log }_{10}x+{\log }_{10}{x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\log }_{10}{x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}---\infty$

$=\left(1+\frac{1}{3}+\frac{1}{9}---\infty \right){\log }_{10}x$

$=\left[\frac{1}{1-{\displaystyle \frac{1}{3}}}\right]{\log }_{10}x=9$

So, $\left(x, y\right)=\left({10}^6, 9\right)$