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five consecutive terms of an arithmetic sequence have a sum of 40. The product of the middle and the two end terms is 224. Find the terms of the sequence.
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5/2(2a+4d)=40 (a+2d)(a+3d)(a+4d)=224 solve for a and d
Lets say the middle term is a and the difference between two terms is d. Than the first term is a-2d the 2nd a-d the 3rd a 4th a+d 5th a+2d the sum is 5a=40 a=8 the product of 1st 3rd 5th is (8-2d)*8*(8+2d)=224 64-4d^2=28 4d^2=36 d=3 or d=-3 So the terms are 2, 5, 8, 11, 14 or 14, 11, 8, 5, 2
the numbers are 2,5,8,11,14
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