Question

please find the answer​

Answer 1

$\begin{aligned}&5^1\times 5^3\times 5^5\times \dots\dots\dots\ \times5^{2n-1}=(25)^{72}\\ \\ \implies\ \ &5^{1+3+5+ \dots\dots\dots\ +2n-1}=(5^2)^{72}\end{aligned}$

Consider the exponent of 5 in the LHS here. It shows an arithmetic series here.

We apply  $S_n=\dfrac{n}{2}[\ 2a+(n-1)d\ ].$

Here,  $a=1\quad;\quad d=3-1=2\quad\quad [\text{or coefficient of n in 2n-1 .}]$

So,

$S_n=\dfrac{n}{2}[\ 2\times 1+(n-1)2\ ]\ =\ n[\ 1+n-1\ ]\ =\ n^2$

Thus,

$\begin{aligned}&5^{1+3+5+ \dots\dots\dots\ +2n-1}=(5^2)^{72}\\ \\ \implies\ \ &5^{(n^2)}=5^{144}\\ \\ \implies\ \ &n^2=144\quad\quad\implies\quad\quad n=\pm 12\end{aligned}$

But since  $n>0,\quad\quad n=\mathbf{12}$