Question

If ${\left(5\right)}^2{\left(5\right)}^4{\left(5\right)}^6\dots .{\left(5\right)}^{2n}={\left(0.04\right)}^{-28}$ , then the value of $n$ is equal to

• A. $7$
• B. $5$
• C. $6$
• D. $3$

Answer 1

Answer: (A)

The given equation can be written as $5^2{\left(5^2\right)}^2{\left(5^2\right)}^3\dots {\left(5^2\right)}^n={\left(5^2\right)}^{28}$
${\left(5^2\right)}^{1+2+3+\dots +n}={\left(5^2\right)}^{28}$
$\Rightarrow \frac{n\left(n+1\right)}{2}=28$
$\Rightarrow n=7$