Question

Find four numbers in AP whose sum is 20 and the sum of whose square 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

a^2 – 6ad + 9d^2 + a^2 – 2ad + d^2 + a^2 + 2ad + d2 + a2 + 6ad + 9d^2 = 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

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