Question

Find four numbers in AP whose sum is 28 and the sum of whose squares is216

Answer 1

Let the four numbers be (a-3d), (a-d), (a+d) & (a+3d).
(Observe that they are in Arithmetic Progression)
(a-3d)+(a-d)+(a+d)+(a+3d)=28 → a=7.
(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216 → 4(a²)+20d²=216 → d=±1.
Hence, the numbers are 4, 6, 8 & 10.

Answer 2

Let the four numbers be (a-3d), (a-d), (a+d) & (a+3d).

Now A.T.Q

1. Their sum
(a-3d)+(a-d)+(a+d)+(a+3d)=28
4a = 28
a=7.

2.Their product
(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216
4a²+20d²=216
a² + 5d² = 54
Putting the value of a
49 + 5d² = 54
5d² = 5
d² = 1
d=±1.

Hence, the numbers are 4, 6, 8 & 10.