Question

If $\frac{1+3+5+\dots \text{upto}n\text{terms}}{4+7+10+\dots \text{upto}n\text{terms}}=\frac{20}{67{\log }_{10}x}$ $\frac{20}{7{\log }_{10}x}$ and $n={\log }_{10}x+{\log }_{10}{x}^{1/2}+{\log }_{10}{x}^{1/4}+\dots ..\infty ,$ then $x$ is equal to

• A. ${10}^3$
• B. ${10}^5$
• C. ${10}^6$
• D. ${10}^7$

Answer 1

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

Given,

$\frac{1+3+5+7\cdot \dots \text{upto nterms}}{4+7+10+-\text{upto n terms}}=\frac{20}{7{\log }_{10}x}$

$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1)2}{\frac{n}{2}[2a_2+(n-1)3}]$

$=\frac{20}{7{\log }_{10}x}$

$\Rightarrow \frac{2+(n-1)2}{2\ast 4+(n-1)3}=\frac{20}{7{\log }_{10}x}$

$\Rightarrow \frac{2_n}{8+3n-3}=\frac{20}{7{\log }_{10}^x}$

$\Rightarrow \frac{n}{5+3n}=\frac{10}{7{\log }^{x}10}-\left(i\right)$

Now,

$n={\log }_{10}^x+{\log }_{10}{x}^{1/2}+{\log }_{10}{x}^{1/4}+\dots \dots \infty$

$\therefore \frac{\log (a{)}^n}{b}=n\frac{\log a}{b}$

$n={\log }_{10}^x+\frac{1}{2}{\log }_{10}^x+\frac{1}{4}{\log }_{10}^x+\cdots \infty$

$m={\log }_{10}^x[1+\frac{1}{2}+\frac{1}{4}+\cdots \cdot \infty )$

$n={\log }_{10}^x\left[\frac{1}{1-\frac{1}{2}}\right]$

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

$n=2{\log }_{10}x$

Now, putting the value n in equation (i)

$\frac{x{\log }_{10}x}{5+3\cdot 2{\log }_{10}x}=\frac{105}{7{\log }_{10}x}$

$\Rightarrow 7\cdot {\left({\log }_{10}x\right)}^2=5(5+6{\log }_{10}^x)$

$\Rightarrow 7{\left({\log }_{10}x\right)}^2=25+30{\log }_{10}^x$

Let ${\log }_{10}^x=t$

$\Rightarrow 7t^2-30t-25=0$

$\Rightarrow 7t^2+5t-35t-25=0$

$\Rightarrow x7t^2-35t+5t-25=0$

$\Rightarrow 7t(t-5)+5(t-5)=0$

$\Rightarrow (7t+5)(t-5)=0$

$\Rightarrow t=-\frac{5}{7}$ or 5

$\because {\log }_{10}x$ does not equal to $-\frac{5}{7}$.

$\begin{array}{cc}\therefore ={\log }_{10}x=5& \\ x=(10{)}^5=\text{option}B& \end{array}$