Q. 135.0( 3 Votes )

# Four numbers are in arithmetic progression. If their sum is 20 and the sum of their squares is 120, then find the numbers.

Answer :

Let the first term be a – 3d and common difference be 2d.

Thus arithmetic progression will be a - 3d, a - d, a + d, a+3d.

Given, Sum of terms = 20

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

⇒ 4a = 20

⇒ a = 5 …(i)

Also,

Sum of their square = 120

(a – 3d)^{2} + (a – d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 120

⇒ (5 – 3d)^{2} + (5 – d)^{2} + (5 + d)^{2} + (5 + 3d)^{2} = 120

⇒ (25 + 9d^{2} – 30d) + (25 + d^{2} – 10d) + (25 + d^{2} + 10d) + (25 + 9d^{2} + 30d) = 120

⇒ 100 + 20d^{2} = 120

⇒ 20d^{2} = 20

⇒ d^{2} = 1

⇒ d = ± 1

If d = 1

First term, a – 3d = 5 – 3 = 2 and common difference, 2d = 2 and four terms will be 2, 4, 6, 8

If d = -1

First term, a – 3d = 5 –(– 3) = 8 and common difference, 2d = -2 and four terms will be 8, 6, 4 and 2.

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