Q.Find the sum of the first $5$ terms of the Geometric Progression $5, 15, 45, \ldots$

Use $S_n = \frac{a(r^n - 1)}{r - 1}$ (convenient since $r > 1$).

Here $a = 5$, $r = \dfrac{15}{5} = 3$ (checked: $\dfrac{45}{15} = 3$), $n = 5$. Since $r = 3 > 1$, use $S_n = \dfrac{a(r^n - 1)}{r - 1}$:

$S_5 = \frac{5(3^5 - 1)}{3 - 1} = \frac{5(243 - 1)}{2} = \frac{5(242)}{2} = \frac{1210}{2} = 605.$ …

Mis-computing $3^5$ as $125$ (that is $5^3$) instead of $243$, or forgetting the $-1$ in the denominator $r - 1$. Alwa …