Question

find the four members in AP so that their sum is 20 and the sum of the sequences 120

Answer 1

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 + 3d)² = 120

⟹ 4a² + 20d² = 120

⟹ a² + 5d² = 30

⟹ 25 + 5d² = 30

⟹ 5d² =5   ⟹ d = ± 1

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

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

Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2

Hope it helps!

Brainliest please! :)