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  4. If (5)2(5)4(5)6 ldots .(5)2 n=(0.04)- 28 , then the value of n is equal to

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

NTA AbhyasNTA Abhyas 2022

Solution:

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