$t_n = a\,r^{\,n-1}$ — read aloud: t-n equals a, times r to the n minus one. Same skeleton as the arithmetic version, same off-by-one, one upgrade: the steps are multiplications now, so they stack as an exponent. The 8th term of 3, 6, 12, … is $3 \times 2^{7} = 384$ — seven doublings, not eight.

Going backwards divides: $a = t_n / r^{\,n-1}$. Recovering the ratio takes a root: $r = (t_n/a)^{1/(n-1)}$, because $t_n/a$ is the growth compounded over ALL the steps, and the root shares it back out one step at a time. This lesson keeps $r$ friendly — 2 and 3 — so the powers stay mental: know your 2s to $2^{9}$ and your 3s to $3^{5}$, and it has no secrets.