Question

The sum of four numbers is 20 and the sum of their squares is 120

Answer 1

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Let the four numbers in A.P be a-3d, a-d,a+d,a+3d.    ---- (1)

Given that Sum of the terms = 20.

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

4a = 20

  a = 5.    ---- (2)

Given that sum of squares of the term = 120.

= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120

= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120

= 4a^2 + 20d^2 = 120

Substitute a = 5 from (2) .

4(5)^2 + 20d^2 = 120

100 + 20d^2 = 120

20d^2 = 20

d = +1 (or) - 1.

Since AP cannot be negative.

Substitute a = 5 and d = 1 in (1), we get

a - 3d, a-d, a+d, a+3d = 2,4,6,8

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Answer 2

Step-by-step explanation:

Let the numbers be (a−3d),(a−d),(a+d),(a+3d).

Then, Sum of numbers =20

⟹(a−3d)+(a−d)+(a+d)+(a+3d)=20⟹4a=20⟹a=5

It is given that, sum of the squares =120

⟹(a−3d)

2

+(a−d)

2

+(a+d)

2

+(a+a+3d)

2

=120

⟹4a

2

+20d

2

=120

⟹a

2

+5d

2

=30

⟹25+5d

2

=30

⟹5d

2

=5⟹d=±1If d=1, the, the numbers are 2,4,6,8.

If d=−1, then the numbers are 8,6,4,2.

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