Problem 11.8. Given an increasing sequence of 8 real numbers. Diana wrote down all possible sequences of 4 consecutive numbers from it. It turned out that two of the five new sequences are arithmetic progressions with differences of 4 and 36, respectively, and one of the sequences is a geometric progression. Find the largest of the given 8 numbers. List all possible options.
Answer: 126, 6.

Solution. Suppose the arithmetic and geometric progressions in the condition can intersect in at least three consecutive elements. Let there be three numbers $a, b$, $c$ that form both an arithmetic and a geometric progression. On one hand, this is an arithmetic progression, so $b-a=c-b$, i.e., $2b=a+c$. On the other hand, $\frac{b}{a}=\frac{c}{b}$, i.e., $b^2=ac$. Therefore, $0=(2b)^2-4b^2=(a+c)^2-4ac=(a-c)^2$, i.e., $a=c$. Since $2b=a+c=2a$, then $a=b=c$. But by the condition, the difference of each arithmetic progression is not equal to 0. Contradiction.

Therefore, the arithmetic and geometric progressions can intersect in no more than two elements. Also, arithmetic progressions with different differences cannot intersect with each other in more than one element in a row. Thus, the only possible scenario is when the first four and the last four numbers form an arithmetic progression, and the middle four numbers form a geometric progression. Consider two cases depending on which of the arithmetic progressions has a difference of 4.

Case 1. Suppose the first 4 numbers form an arithmetic progression with a difference of 4. Let the third number be $a$, then the fourth is $a+4$. Let the fifth number be $b$, then the sixth is $b+36$. These four numbers form a geometric progression if and only if $\frac{a+4}{a}=\frac{b}{a+4}=\frac{b+36}{b}$. By equating the first and third fractions, it is easy to derive that $b=9a$. Substituting $b=9a$ into the first equation, we get the quadratic equation $8a^2-8a-16=0$ with roots $a_1=-1$ and $a_2=2$. In the first case, we get the sequence $-9,-5,-1,3,-9,27,63,99$, but it is not increasing. In the second case, we get the sequence $-6,-2,2,6,18,54,90,126$, and this one fits.

Case 2. Suppose the first 4 numbers form an arithmetic progression with a difference of 36. Similarly to the previous case, we get that $\frac{a+36}{a}=\frac{b}{a+36}=\frac{b+4}{b}$. As in the previous case, it is not difficult to derive that $a=9b$, and then obtain the quadratic equation $8b^2+72b+144=0$ with roots $b_1=-6$ and $b_2=-3$. In the first case, we get the sequence $-126,-90,-54,-18,-6,-2,2,6$, and it fits. In the second case, we get the sequence $-99,-63,-27,9,-3,1,5,9$, but it is not increasing.

Thus, there are two possible sequences. The largest numbers in them are 126 and 6, respectively.