Q.The population of a town is $10{,}000$ and grows at a constant rate of $10\%$ per year. Treating the present population as the 1st term of a Geometric Progression, find the population after $3$ years.

A $10\%$ annual growth means each year's population is $1.1$ times the previous — a GP with $r = 1.1$; "after 3 years" is the 4th term.

A constant percentage growth each year gives a GP: $a = 10000$ (present population) and $r = 1.1$ (each year's population is $110\%$, i.e. $1.1$ times, the previous year's). With the present population as the 1st term, the population after 3 years is the 4th term ($n = 4$).

Apply $t_n = a\,r^{n-1}$:

Compute $(1.1)^3$: $(1.1)^2 = 1.21$, then $1.21 \times 1.1 = 1.331$. So $t_4 = 10000 \times 1.331 = 13310$. …

Miscounting the exponent: "after 3 years" means $3$ years of growth from the present, so the exponent on $r$ is $3$ (i.e. $n - 1 = 3$, giving the 4th term …