The sum of the four consecutive number an AP is 37 and the ratio of the product of the first and the last term of the product of to middle term is 7:15 find?
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Let the consecutive number be
=a-3d a-d a+d a+3d
=Sum is given as 37
=4a=37 a=37/4
Given ratio of products is 7:15
=(a-3d) (a+3d)/(a-d) (a+d) = 7/15
=15×(a^2-9d^2)= 7×(a^2-d^2)
=a^2 = 16d^2
=d=a/4 or -a/4
=d= 37/16 or -37/16
Therefore,the numbers ( by taking positive value for d)are 37/16,111/16,185/16,259/16.
NOTE:AP may be formed by reversing the number also (by taking negative value for d)
HOPE THIS HELPS....
Let the consecutive number be
=a-3d a-d a+d a+3d
=Sum is given as 37
=4a=37 a=37/4
Given ratio of products is 7:15
=(a-3d) (a+3d)/(a-d) (a+d) = 7/15
=15×(a^2-9d^2)= 7×(a^2-d^2)
=a^2 = 16d^2
=d=a/4 or -a/4
=d= 37/16 or -37/16
Therefore,the numbers ( by taking positive value for d)are 37/16,111/16,185/16,259/16.
NOTE:AP may be formed by reversing the number also (by taking negative value for d)
HOPE THIS HELPS....
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