Four numbers are in A.P. If their sum is 20 and sum of their squares

1.	Four numbers are in A.P. If their sum is 20 and sum of their squares is 120, then find the numbers.
Answer» Let in four numbers, First number = a – 3d Second number = a – d Third number = a + d Fourth number = a + 3d Sum of four number is 20 ∴ 20 = (a – 3d) + (a – d) + (a + d) + (a + 3d) ⇒ 20 = 4a ⇒ a = 5 Sum of squares of four numbers is 120. ∴ (a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 120 ⇒ [a² + 9d² – 6ad + a² + d² – 2ad + a² + d² + 2ad + a² + 9d² + 6ad] = 120 ⇒ 4 (a² + 5d²) = 120 ⇒ a² + 5d² = 30 [∵ a = 5] ⇒ 5² + 5d² = 30 ⇒ 25 + 5d² = 30 ⇒ 5d² = 30 – 25 ⇒ 5d² = 5 ⇒ d² = 1 ⇒ d = ±1 Thus, a = 5 and d = ± 1 ∴ Four numbers are 2, 4, 6, 8 or 8, 6, 4, 2

Four numbers are in A.P. If their sum is 20 and sum of their squares is 120, then find the numbers.

Answer»

Let in four numbers,

First number = a – 3d

Second number = a – d

Third number = a + d

Fourth number = a + 3d

Sum of four number is 20

∴ 20 = (a – 3d) + (a – d) + (a + d) + (a + 3d)

⇒ 20 = 4a

⇒ a = 5

Sum of squares of four numbers is 120.

∴ (a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 120

⇒ [a² + 9d² – 6ad + a² + d² – 2ad + a² + d² + 2ad + a² + 9d² + 6ad] = 120

⇒ 4 (a² + 5d²) = 120

⇒ a² + 5d² = 30 [∵ a = 5]

⇒ 5² + 5d² = 30

⇒ 25 + 5d² = 30

⇒ 5d² = 30 – 25

⇒ 5d² = 5

⇒ d² = 1

⇒ d = ±1

Thus, a = 5 and d = ± 1

∴ Four numbers are 2, 4, 6, 8 or 8, 6, 4, 2