## Problem 3

Determine the sum of the first 30 terms of a geometric progression, knowing that the sum of the first two terms is equal to 8, and the difference between the second term and the first term is equal to 4.
## Problem 3

$\left\{\begin{array}{l}b_{1}+b_{2}=8 \\ b_{2}-b_{1}=4\end{array} \Leftrightarrow\left\{\begin{array}{l}b_{1}(1+q)=8 \\ b_{1}(q-1)=4\end{array}\right.\right.$

By dividing the relations member by member $\Rightarrow q=3$, then $b_{1}=2$ $3 p$

$S_{30}=\frac{b_{1}\left(q^{30}-1\right)}{q-1}=3^{30}-1$.