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## Answers

(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.

The Brainliest Answer!

*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.*