Three consecutive terms of an arithmetic sequence have a sum of 12 and a product of -80 .Find the terms. [hint : let the terms be x - d, x and x + d . ]
so, follow the hint. add them and multiply them.
56, 49
To solve this problem, we can use the given hint and express the three consecutive terms of the arithmetic sequence as x-d, x, and x+d.
Given that the sum of these three terms is 12, we can write the equation:
(x - d) + x + (x + d) = 12
Simplifying the equation, we get:
3x = 12
Dividing both sides by 3, we find:
x = 4
Now, let's focus on the second given condition, which states that the product of the three terms is -80. We can write this as an equation:
(x - d) * x * (x + d) = -80
Substituting the value of x we found earlier (x = 4), we have:
(4 - d) * 4 * (4 + d) = -80
Expanding the equation, we get:
(16 - 4d) * (4 + d) = -80
Now, we can solve this equation for d.
16(4 + d) - 4d(4 + d) = -80
64 + 16d - 16d - 4d² = -80
Rearranging terms, we get:
4d² - 16d + 144 = 0
Dividing both sides by 4, we have:
d² - 4d + 36 = 0
This quadratic equation does not factorize easily, so we can use the quadratic formula:
d = ( -b ± √(b² - 4ac) ) / 2a
In this case, a = 1, b = -4, and c = 36. Substituting these values in, we get:
d = ( -(-4) ± √((-4)² - 4(1)(36)) ) / (2 * 1)
Simplifying further, we have:
d = (4 ± √(16 - 144)) / 2
d = (4 ± √(-128)) / 2
We encounter a problem here because the square root of a negative number is not a real number. This means there is no real value for d that satisfies the equation.
Therefore, there are no three consecutive terms in the arithmetic sequence that have a sum of 12 and a product of -80.