Question

$\frac{1+3+5+...+n}{4+7+10+.....+n}$=$\frac{20}{7{\log }_{10}x}$ and n = ${\log }_{10}x+{\log }_{10}{x}^{1/2}+{\log }_{10}{x}^{1/4}+....$, then x is equal to

• A. ${10}^3$
• B. ${10}^5$
• C. ${10}^7$
• D. ${10}^9$

Answer 1

The correct option is B ${10}^5$

$n=\log x+\frac{1}{2}\log x+\frac{1}{4}\log x+.........$

$n=\left(\log x\right)(1+\frac{1}{2}+.....)$

$n=\left(\log x\right)\left(\frac{1}{1-\frac{1}{2}}\right)=2lo{g}_{10}x$

Now, in numerator,

$1+3+.....\left(nterms\right)={\sum }_{k=1}^n(2k-1)=2\left(\frac{n(n+1)}{2}\right)-n$

$4+7+10+......\left(nterms\right)={\sum }_{k=1}^n(3k+1)=\frac{3n(n+1)}{2}+n$

Now,

$\frac{n^2}{\frac{n}{2}(3n+5)}=\frac{20}{7{\log }_{10}x}$

$\frac{2n}{3n+5}=\frac{20}{7{\log }_{10}x}$

$let{\log }_{10}x=z$

$\therefore n=2z$

$28z^2=20(6z+5)$

$7z^2-30z-25=0$

$\therefore z=5$

${\log }_{10}x=5$

$x={10}^5$