Solve the equation: $\;\;\;$ $5^2 \times 5^4 \times 5^6 \times \cdots \times 5^{2x} = \left(0.04\right)^{-28}$
Given equation: $\;\;\;$ $5^2 \times 5^4 \times 5^6 \times \cdots \times 5^{2x} = \left(0.04\right)^{-28}$
i.e. $\;$ $5^{\left[2 + 4 + 6 + \cdots + 2x\right]} = \left(\dfrac{1}{25}\right)^{-28}$
i.e. $\;$ $5^{\left[2 + 4 + 6 + \cdots + 2x\right]} = \left(5^{-2}\right)^{-28} = 5^{56}$
$\implies$ $2 + 4 + 6 + \cdots + 2x = 56$ $\;\;\; \cdots \; (1)$
Now, $\;$ $2 + 4 + 6 + \cdots + 2x$ $\;$ is an arithmetic progression (A.P) with
first term $= a = 2$,
common difference $= d = 2$,
$n^{th}$ term $= t_n = 2x$ $\;$ and
sum to $n$ terms $= S_n = 56$
$n^{th}$ term of an A.P $= t_n = a + \left(n - 1\right)d$
i.e. $\;$ $2x = 2 + \left(n - 1\right) \times 2$
i.e. $\;$ $x = 1 + n - 1$ $\implies$ $x = n$ $\;\;\; \cdots \; (2)$
Sum to $n$ terms of an A.P $= S_n = \dfrac{n}{2} \times \left[2a + \left(n - 1\right)d\right]$
i.e. $\;$ $56 = \dfrac{n}{2} \times \left[2 \times 2 + \left(n - 1\right) \times 2\right]$
i.e. $\;$ $112 = n \left[4 + 2n - 2\right]$
i.e. $\;$ $2n^2 + 2n - 112 = 0$
i.e. $\;$ $n^2 + n - 56 = 0$
i.e. $\;$ $\left(n - 7\right) \left(n + 8\right) = 0$
i.e. $\;$ $n - 7 = 0$ $\;$ or $\;$ $n + 8 = 0$
i.e. $\;$ $n = 7$ $\;$ or $\;$ $n = -8$
Since the number of terms of an A.P cannot be negative,
$\implies$ $n = 7$
$\therefore \;$ By equation $(2)$, $\;$ $x = 7$