Find four numbers in A.P such that their sum is 16 and their product is 105.

Let the four number be (a – 3d), (a – d), (a + d), (a + 3d)

Given: Sum of numbers = 16

⇒ a – 3d + a – d + a + d + a + 3d = 16

⇒ 4a = 16

and product of numbers = 105

⇒(a – 3d), (a + 3d), (a – d), (a + d) = 105

⇒(a² – 9d²), (a² + d²) = 105

⇒(a²)² 10 a²d² + 9 (d²)² = 105

⇒(4⁴) –10 × 4⁴ × d² + 9d⁴ = 105

⇒9d⁴ – 160 d² + 256 – 105 = 0

⇒9d⁴ – 9d² – 151d² + 151 = 0

⇒9d²(d²– 1) – 151(d² – 1) = 0

⇒(9d²– 151) (d² – 1) = 0

Since a and d are integers

Thus d = ± 1

Hence the four numbers are (4 – 3), (4 – 1), (4 +1), (4 + 3) = 1, 3, 5, 7