please solve this question
Attachments:
Answers
Answered by
2
let the four consecutive numbers be a-3d,a-d,a+d,a+3d.
a-3d+a-d+a+d+a+3d=32
4a=32
a=8
(a-3d)(a+3d)/(a-d)(a+d)=7/15
a²-9d²/a²-d²=7/15
15a²-135d²=7a²-7d²
8a²=128d²
a²=16d²
a=4d
8=4d
d=2
Therefore, numbers are
8-3(2),8-2,8+2,8+3(2)
8-6,8-2,8+2,8+6
2,6,10,14
Hope it helps ! Cheers
a-3d+a-d+a+d+a+3d=32
4a=32
a=8
(a-3d)(a+3d)/(a-d)(a+d)=7/15
a²-9d²/a²-d²=7/15
15a²-135d²=7a²-7d²
8a²=128d²
a²=16d²
a=4d
8=4d
d=2
Therefore, numbers are
8-3(2),8-2,8+2,8+3(2)
8-6,8-2,8+2,8+6
2,6,10,14
Hope it helps ! Cheers
Similar questions