Question

Find four numbers in AP whose sum is 28 and the sum of squares is 216
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Answer 1

Answer:

4, 6, 8 & 10.

Step-by-step explanation:

(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

THERE ARE 2 DIFFRENT WAYS -

1) 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.

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.

2) 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.