Question

Find 4 number in ap whose sum is 20 and the sum of tbose square is 120

Answer 1

Step-by-step explanation:

Given Find 4 number in ap whose sum is 20 and the sum of whose square is 120

• consider the numbers to be a – 3d, a – d, a + d, a + 3d
• Sum of numbers = a – 3d + a – d + a + d + a + 3d = 20
•                           4a = 20
•                       Or a = 5
• According to question Sum of the squares of the numbers will be
• (a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 20
• So a^2 – 6ad + 9d^2 + a^2 – 2ad + d^2 + a^2 + 2ad + d^2 + a^2 + 6ad + 9d^2 = 120
•  4a^2 + 20d^2 = 120------------------1
• Substituting a = 5 in 1 we get
• 4(5)^2 + 20 d^2 = 120
• 100 + 20 d^2 = 120
• 20d^2 = 20
• Or d^2 = 1
• Or d = + - 1
• Therefore the four numbers are by taking d = 1
• (5 – 3),(5 – 1),(5 + 1),(5 + 3)
• = 2, 4, 6, 8
• Also if d = - 1 we get
• (5 + 3),(5 + 1),(5 – 1),(5 – 3)
• = 8,6,4,2

Reference link will be

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Answer 2

The terms in the A.P. are 2, 4, 6, 8 or 8, 6, 4, 2.

Step-by-step explanation:

Let the terms in the A.P. are (a - 3d), (a - d), (a + d) and (a + 3d).

Now, the sum of them is 20.

So, (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20

⇒ a = 5 ........... (1)

Now, sum of their squares is 120.

So, (a - 3d)² + (a + 3d)² + (a - d)² + (a + d)² = 120

⇒ 2(a² + 9d²) + 2(a² + d²) = 120

⇒ 2a² + 10d² = 60

⇒ 50 + 10d² = 60 {Since a = 5}

⇒ d² = 1

⇒ d = ± 1

Therefore, the terms in the A.P. are 2, 4, 6, 8 or 8, 6, 4, 2. (Answer)