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tiwaribianca475
26.06.2020 •
Mathematics
The sum of four consecutive numbers in an A.P. is 28. The product of the
second and third numbers exceeds that of the first and last by 18. Find the
numbers.
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Ответ:
Let x be the first number in this sequence of four. Then the next three are x + n, x + 2n, and x + 3n, because consecutive terms in an arithmetic progression differ by a fixed constant.
The sum of these four numbers is 28:
x + (x + n) + (x + 2n) + (x + 3n) = 4x + 6n = 28
or
2x + 3n = 14
The product of the second and third numbers is greater than the product of the first and last numbers by 18:
(x + n)(x + 2n) = x(x + 3n) + 18
Expanding everything on both sides gives
x² + 3nx + 2n² = x² + 3nx + 18
and terms cancel so we end up with
2n² = 18 ==> n² = 9 ==> n = 3 or n = -3
If n = 3, then
4x + 6*3 = 28 ==> 4x = 10 ==> x = 5/2
If n = -3, then
4x + 6*(-3) = 28 ==> 4x = 46 ==> x = 23/2
So we have two possible solutions, but the only difference between them is the order. If n = 3, then the four numbers are 5/2, 11/2, 17/2, and 23/2. If n = -3, then the four numbers are, again, 23/2, 17/2, 11/2, and 5/2.
Ответ:
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