Sosukhin 0.

A sequence of two different numbers was continued in two ways: to form a geometric progression and to form an arithmetic progression. The third term of the geometric progression coincided with the tenth term of the arithmetic progression. With which term of the arithmetic progression did the fourth term of the geometric progression coincide?
Let $a$ be the first of two numbers in the original sequence, $d$ be the difference of the arithmetic progression, and $q$ be the ratio of the geometric progression. Then, according to the problem, $a+d=a q, a+9 d=a q^{2}$. Therefore, $a(q-1)=d$ and $a(q-1)(q+1)=a\left(q^{2}-1\right)=9 d=9 a(q-1)$. Since $q \neq 1$, from this we get $q=8$ and $a q^{3}=a+a\left(q^{3}-1\right)=$ $a+a(q-1)\left(q^{2}+q+1\right)=a+73 d$. Thus, the fourth term of the geometric progression coincides with the 74th term of the arithmetic progression.

## Answer

with the 74th term of the arithmetic progression.