Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 3
\(\Rightarrow\) 3a = 3
\(\Rightarrow\) a = 1
Also, (a - d).a.(a + d) = -35
\(\Rightarrow\) a(a2 - d2) = -35
\(\Rightarrow\) 1.(1 - d2) = -35
\(\Rightarrow\) d2 = 36
\(\Rightarrow\) d = \(\pm6\)
Thus, a = 1 and d = \(\pm6\)
Hence, the required numbers are (-5,1 and 7) or (7,1 and -5).
