17. Solution. Let the first of the six addends be \(a\), and the second be \(b\). Then the third, fourth, fifth, and sixth addends will be \(a+b\), \(a+2b\), \(2a+3b\), and \(3a+5b\) respectively. The sum of all six numbers is

\[
\begin{gathered}
a+b+(a+b)+(a+2b)+(2a+3b)+(3a+5b)= \\
= \ldots = 4(2a+3b)
\end{gathered}
\]

i.e., it is equal to four times the fifth addend. Therefore, if the sum of the six numbers is known, dividing it by 4 will give the fifth number.
17*. Vovochka wrote two numbers on the board. The third one he wrote as the sum of the first two, the fourth as the sum of the third and the second, and so on. Then Vovochka told Sasha the sum of six consecutive numbers, starting from one of the written ones. Upon learning the sum, Sasha immediately determined one of the written numbers. Which one?

18. Among the polynomials:

\[
\begin{aligned}
& -2a-1, \quad 2a-1, \quad -a+1, \quad a+1 \\
& -2a-2, \quad 2a-2, \quad -a+2, \quad a+2 \\
& -2a-3, \quad 2a-3, \quad -a+3, \quad a+3 \\
& -2a-4, \quad 2a-4, \quad -a+4, \quad a+4
\end{aligned}
\]
- are there two polynomials whose sum is \(a - 3\)? How many such pairs of polynomials are there? How many pairs of polynomials are there whose difference is \(a + 4\)?