Find four numbers in A.P., whose sum is 20 and sum of whose squares is 120.

Asked by Amuthavalli Thayyal Gopala Krishnan | 23rd Nov, 2013, 08:38: PM

Expert Answer:

Let the terms ba – 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

Answered by  | 23rd Nov, 2013, 11:27: PM

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