If we add the same number to 5, 7, and 10, we get the squares of three consecutive terms of an arithmetic sequence. What is the difference of the sequence, and what number did we add to the numbers?
I. solution. Let the added number be $x$, then according to the known property of three consecutive terms of an arithmetic sequence,

$$
\sqrt{5+x}+\sqrt{10+x}=2 \sqrt{7+x}
$$

By squaring and rearranging,

$$
2 \sqrt{(5+x)(10+x)}=(28+4 x)-(15+2 x)=13+2 x
$$

Squaring again, $x=-31 / 8$, no other number can satisfy the requirement. This indeed satisfies the condition if we take all three terms of the sequence to be positive:

$$
\frac{3}{2 \sqrt{2}}, \quad \frac{5}{2 \sqrt{2}}, \quad \frac{7}{2 \sqrt{2}}
$$

and also if we take all three to be negative. In the first case, $1 / \sqrt{2}$ is the difference of the sequence, in the second case, $-1 / \sqrt{2}$.

Imre Gulyás (Budapest, Piarist g. II. grade)

Cecilia Takács (Veszprém, Lovassy L. g. III. grade)

II. solution. Let the middle term of the sequence be $a$, the difference be $d$, and the added number again be $x$; for these, the following holds

$$
5+x=(a-d)^{2}, \quad 7+x=a^{2}, \quad 10+x=(a+d)^{2}
$$

Subtracting twice the second equation from the sum of the first and third equations,

$$
1=2 d^{2}, \quad d= \pm 1 / \sqrt{2}= \pm \sqrt{0.5}
$$

Subtracting the first equation from the second,

$$
2=2 a d-d^{2}, \quad a=\frac{2+d^{2}}{2 d}=\frac{5}{4 d}= \pm \frac{5}{2 \sqrt{2}}= \pm \sqrt{3.125}
$$

Finally, from the second equation,

$$
x=a^{2}-7=-\frac{31}{8}
$$

According to this, two sequences satisfy the conditions, the middle term of the sequence with a difference of $\sqrt{0.5}$ is $\sqrt{3.125}$, and the middle term of the sequence with a difference of $-\sqrt{0.5}$ is $-\sqrt{3.125}$, but the added term in both cases is $-31 / 8$.

Gabriella Gádoros (Tatabánya, Árpád g. III. grade)

László Váli (Budapest, I. István g. I. grade)