Sum of three numbers in A.P. is 3 and their product is — 35. Find the numbers.

Let the three numbers are (a – d), a and (a + d)

According to question,

Sum of these three numbers = 3

⇒ a – d + a + a + d = 3

⇒ 3a = 3

⇒ a = 1 …(i)

and it is also given that

Product of these numbers = – 35

⇒(a – d) × a × (a + d) = – 35

⇒(1 – d) × 1 × (1 + d) = – 35

⇒ 1 × (1² – d²) = – 35 [∵ (a – b)(a + b) = a² – b²]

⇒ 1 × (1 – d²) = – 35

⇒ 1 – d² = – 35

⇒ – d² = – 35 – 1

⇒ – d² = – 36

⇒ d² = 36

⇒ d = √36

⇒ d = ±6

Case I: If d = 6 and a = 1

a – d = 1 – 6 = – 5

a = 1

a + d = 1 + 6 = 7

So, the numbers are

– 5, 1, 7

Case II: If d = – 6 and a = 1

a – d = 1 – ( – 6) = 1 + 6 = 7

a = 1

a + d = 1 + ( – 6) = 1 – 6 = – 5

So, the numbers are

7, 1, – 5