3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,14, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points $)$
Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula

$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$

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 $5+3 n=10 \cdot 2^{k}$ must hold, from which

$$
n=\frac{10 \cdot 2^{k}-5}{3}=\frac{5 \cdot\left(2^{k+1}-1\right)}{3}, \text { i.e., }\left(2^{k+1}-1\right): 3
$$

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

For $k=1, n=5$ the common element is $a_{5}=b_{1}=20$;

for $k=3, n=25$ the common element is $a_{25}=b_{3}=80$;
for $k=5, n=105$ the common element is $a_{105}=b_{5}=320$ and so on.

Thus, the required elements are: $20,80,320, \ldots$ - 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 20. The sum of the first 10 terms can be found using the formula $S=c_{1} \frac{q^{n}-1}{q-1}=20 \cdot \frac{4^{10}-1}{4-1}=20 \cdot \frac{\left(2^{10}\right)^{2}-1}{3}=20 \cdot \frac{1024^{2}-1}{3}=6990500$.

Answer. 6990500.