The sequence (tn ) is a G.P.
if \(\frac{t_{n+1}}{t_n}\) = constant for all n ∈ N.
Now, \(t_n=\frac{5^{n-2}}{7^{n-3}}\)

= 5/7 = constant, for all n ∈ N.
\(\therefore\) the sequence is a G.P. with common ration = 5/7
\(\therefore\) first term = t1 = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^2}5=\frac{49}5\)