Recover geometric terms from partial sums

For $n\ge2$, $u_n=S_n-S_{n-1}=\dfrac1a[(a-1)^{n-1}-(a-1)^n]=\dfrac{2-a}a(a-1)^{n-1}$. At $n=1$, $u_1=S_1=(2-a)/a$, so the formula holds for every positive integer $n$.

$u_{n+1}=(a-1)u_n$, so the underlying terms form a GP with common multiplier $a-1$. The printed ratio proof $u_{n+1}/u_n=a-1$ additionally needs $a\ne2$. At $a=2$, all terms and partial sums are zero; the multiplicative relation remains true but the ratio is $0/0$. The question’s wording refers to the underlying geometric series: $S_n$ is its partial sum, not generally itself a geometric progression.