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 $\{10,20,40,80, \ldots\} \cdot(10$ 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 $\{10,20,40,80, \ldots\}$ are given by the formula

$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$

For the common elements, the equality $4+3 n=10 \cdot 2^{k}$ must hold, from which

$$
n=\frac{10 \cdot 2^{k}-4}{3}=\frac{2 \cdot\left(5 \cdot 2^{k}-2\right)}{3}, \text { i.e., }\left(5 \cdot 2^{k}-2\right) \vdots 3 .
$$

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

For $k=0, n=2$ the common element is $a_{2}=b_{0}=10$;

for $k=2, n=12$ the common element is $a_{12}=b_{2}=40$;
for $k=4, n=52$ the common element is $a_{52}=b_{4}=160$ and so on.

Thus, the sought elements are: $10,40,160, \ldots$ - the members of the original geometric progression with even indices. They form a geometric progression with a common ratio of 4 and the first term 10. The sum of the first 10 terms can be found using the formula $S=c_{1} \frac{q^{n}-1}{q-1}=10 \cdot \frac{4^{10}-1}{4-1}=10 \cdot \frac{\left(2^{10}\right)^{2}-1}{3}=10 \cdot \frac{1024^{2}-1}{3}=3495250$.

Answer. 3495250.