If 5^25^45^6….5^2n=0.04^(-28), then the value of n is equal to

Correct answer: (A) 7. The given equation can be written as 5^25^225^23…5^2n=5^228 5^21+2+3+…+n=5^228 ⇒(nn+1)/2=28 ⇒n=7

Q15 / 118 · NTA Abhyas (JEE Main) 2020 · Mathematics · Sequences and Series

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

The given equation can be written as

5^{2} {(5^{2})}^{2} {(5^{2})}^{3} \dots {(5^{2})}^{n} = {(5^{2})}^{28}

{(5^{2})}^{1 + 2 + 3 + \dots + n} = {(5^{2})}^{28}

\Rightarrow \frac{n (n + 1)}{2} = 28

\Rightarrow n = 7

Hence, the correct answer is option: 'A'.