Giả sử cấp số nhân đã cho là: $a,aq,\ldots,aq^{n-1}, q\ne0$
Theo giả thiết ta có: $\left\{\begin{array}{l}\sum_{i=0}^{n-1}aq^i=11\\\sum_{i=0}^{n-1}aq^{2i}=341\\\sum_{i=0}^{n-1}aq^{3i}=3641\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}\dfrac{a(1-q^n)}{1-q}=11\\\dfrac{a(1-q^{2n})}{1-q^2}=341\\\dfrac{a(1-q^{3n})}{1-q^3}=3641\end{array}\right.$
Suy ra: $\left\{\begin{array}{l}\dfrac{1+q^n}{1+q}=31\\\dfrac{1+q^n+q^{2n}}{1+q+q^2}=331\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}q^n=31q+30\\1+31q+30+(31q+30)^2=331(1+q+q^2)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}q^n=31q+30\\21q^2+52q+20=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}q^n=31q+30\\\left[\begin{array}{l}q=-2\\q=-\dfrac{10}{21}\end{array}\right.\end{array}\right.$
Với $q=-2$. suy ra: $q^n=-32 \Leftrightarrow n=5$
Với $q=-\dfrac{10}{21}$. suy ra: $q^n=\dfrac{320}{21}$, loại.
Vậy CSN đã cho có 5 số hạng.