Question

Fond four numbers in AP whose sum is 20 and the sum of whose squares is 120

Answer 1

Let the terms be a – 3d, a – d, a + d, a+3d

Sum of the terms = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

4a = 20

a = 5

Sum of the squares of the term

= (a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 20

a² – 6ad + 9d² + a² – 2ad + d² + a² + 2ad + d² + a² + 6ad + 9d² = 120

4a² + 20d² = 120 – – – – – – – – – – – – (a)

Substituting a = 5 into (a)

4(5²) + 20d² = 120

100 + 20d² = 120

d = + 1

d = + 1

Thus, the four numbers are:

Taking d = 1

(a – 3d), (a – d), (a + d), (a + 3d)

= (5 – 3), (5 – 1), (5 + 1), (5 + 3)

= 2, 4, 6, 8

Taking d = -1

(a – 3d), (a – d), (a + d), (a + 3d)

= (5 + 3), (5 + 1), (5 - 1), (5 - 3)

= 8, 6, 4, 2