The sum of four consecutive numbers of an AP is 32 and the ratio of the product of the 1st and the last term to the product of two middle terms is 7:15. find the numbers
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Answer:
2 , 6 , 10 , 14
or
14 , 10 , 6 , 2
Step-by-step explanation:
let say 4 numbers are
a , a+d , a +2d , a +3d
Sum = 4a + 6d = 32
2a + 3d = 16
a(a+3d) / (a+d)((a+2d) = 7/15
15(a^2 + 3ad) = 7 ( a^2 + 3ad + 2d^2)
15a^2 + 45ad = 7a^2 + 21ad + 14d^2
8a^2 + 24ad - 14d^2 = 0
4a^2 + 12ad - 7d^2 = 0
4a^2 - 2ad + 14ad - 7d^2 = 0
2a(2a -d) + 7d(2a - d) = 0
(2a + 7d)(2a -d) = 0
2a = -7d or d
case 1 2a = -7d
-7d + 3d = 16
-4d = 16
d = -4
2a = -7(-4)
a = 14
numbers are
14 , 10 , 6 , 2
case 2
2a = d
d + 3d = 16
4d = 16
d = 4
a = 4/2 = 2
numbers are
2 , 6 , 10 , 14
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