Q.Find the sum of the first 6 terms of the Geometric Progression $3, 6, 12, \ldots$

Use $S_n=\frac{a(r^n-1)}{r-1}$ with $a=3, r=2, n=6$ (convenient since $r>1$).

Here $a=3$, $r=\dfrac{6}{3}=2$ (checked: $\dfrac{12}{6}=2$ — constant), $n=6$. Since $r=2>1$, use $S_n=\dfrac{a(r^n-1)}{r-1}$.

$S_6=\dfrac{3(2^6-1)}{2-1}=\dfrac{3(64-1)}{1}=3(63)=189$. …

Forgetting to subtract $1$ in the denominator $r-1$ (using just $r$) is a common slip. Also, since $r^6=2^6$ grows quickly, a common arithmetic error is mis- …