3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\} .(10$ points $)$
Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula

$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$

The members of the geometric progression $(20,40,80, \ldots\}$ are given by the formula

$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$

For common elements, the equality $4+3 n=20 \cdot 2^{k}$ must hold, from which $n=\frac{20 \cdot 2^{k}-4}{3}=\frac{4 \cdot\left(5 \cdot 2^{k}-1\right)}{3}$, that is, $\left(5 \cdot 2^{k}-1\right): 3$.

It is clear that $5 \cdot 2^{k}-1$ is divisible by 3 for odd $k$ (the remainder of the division of $2^{k}$ by 3 is 2 or 1).

For $k=1, n=12$ the common element is $a_{12}=b_{1}=40$;
for $k=3, n=52$ the common element is $a_{52}=b_{3}=160$;

for $k=5, n=212$ the common element is $a_{212}=b_{5}=640$ and so on.

Thus, the required elements are: $40,160,640, \ldots$ - the members of the original geometric progression with odd indices. They form a geometric progression with a common ratio of 4 and the first term 40. The sum of the first 10 terms can be found using the formula $S=c_{1} \frac{q^{n}-1}{q-1}=40 \cdot \frac{4^{10}-1}{4-1}=40 \cdot \frac{\left(2^{10}\right)^{2}-1}{3}=40 \cdot \frac{1024^{2}-1}{3}=13981000$.

Answer. 13981000.