Question

Find four numbers in A.P. whose sum is 40 and the sum of whose squares is 480.

Answer 1

Answer:

So, the Arithmetic Progression is 4,8,12,16.

Answer 2

Question :- Find four numbers in A.P. whose sum is 40 and the sum of whose squares is 480. ?

Solution :-

Let us assume that, given four numbers in AP are (a - 3d) , (a - d) , (a + d) , (a + 3d) .

Than,

→ sum of four numbers = 40

→ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 40

→ (a + a + a + a) + (3d + d - 3d - d) = 40

→ 4a + (4d - 4d) = 40

→ 4a = 40

→ a = 10.

Also,

→ Sum of Square of four numbers = 480

→ (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 480 .

using :-

• (a - b)² = a² + b² - 2ab
• (a + b)² = a² + b² + 2ab

→ (a² + 9d² - 6ad) + (a² + d² - 2ad) + (a² + d² + 2ad) + (a² + 9d² + 6ad) = 480

→ (a² + a² + a² + a²) + (9d² + d² + d² + 9d²) + (6ad + 2ad - 6ad - 2ad) = 480

→ 4a² + 20d² = 480

→ 4(a² + 5d²) = 480

→ a² + 5d² = 120

Putting value of a = 10 Now,

→ (10)² + 5d² = 120

→ 100 + 5d² = 120

→ 5d² = 120 - 100

→ 5d² = 20

→ d² = 4

Square - root both sides,

→ d = ±2

Taking , d = +2

Substituting a = 10 and d = 2 , we get,

→ (a - 3d) = 10 - 3*2 = 10 - 6 = 4.

→ (a - d) = 10 - 2 = 8.

→ (a + d) = 10 + 2 = 12.

→ (a + 3d) = 10 + 3*2 = 10 + 6 = 16.

Taking, d = (-2).

Substituting a = 10 and d = (-2) , we get,

→ (a - 3d) = 10 - 3*(-2) = 10 + 6 = 16.

→ (a - d) = 10 - (-2) = 10 + 2 = 12.

→ (a + d) = 10 - 2 = 8.

→ (a + 3d) = 10 + 3*(-2) = 10 - 6 = 4.

Hence, Required Four Numbers are :- (4,8,12,16) or, (16,12,8,4).