Let numbers a, b and c are in AP.
∴ 2b = a + c. … (1)
Given that sum of these numbers is 3.
∴ a + b + c = 3
⇒ 2b + b = 3 (From equation (1), a + c = 2b)
⇒ 3b = 3
⇒ b = 1.
Now putting b = 1, in equation (1), we get
a + c = 2. … (2)
Given that the product of these numbers is –35
∴ abc = –35
⇒ ac = –35. (∵ b =1)
Now ( a − c )2 = ( a + c )2 − 4ac = 22 – (4 × – 35) = 4 + 140 = 144.
(∵ a + c = 2, ac = –35)
⇒ a – c = \(\sqrt{144}\) = 12
⇒ a – c = 12. … (3)
Now, adding equations (2) and (3), we get
(a + c)+(a – c) = 2 +12 = 14
⇒ 2a = 14
⇒ a = \(\frac{14}2\) = 7.
Now, putting a = 7 in equation (2), we get
7 + c = 2 ⇒ c = 2 –7 = –5.
∴ a = 7, b = 1 and c = –5.
Hence, the numbers are 7, 1 and –5.