Question

three consecutive terms of an aritmetic sequence have a sum of 12 and a product of -80. Find the three terms

Answer 1

Let the first term be \(a\) and the common difference be \(d\). Try creating 2 equations from that :)

Answer 2

Hint: the second time is \(a+d\) :)

Answer 3

thanks

Answer 4

Let x=first term and k=common difference;

(x)+(x+k)+(x+2k)=12 \\since it has been said that the sum is 12.
3x+3k=12 \\combining like terms
x+k=4 \\multiplying both sides by 1/3
k=4-x \\this is the value of the common difference

substitute:

(x)(x+k)(x+2k)=-80 \\since it has been said that the product is -80
(x)[(x)+(4-x)][(x)+2(4-x)]=-80 \\substituting the value of k
(x)(4)(8-x)=-80 \\performing the operation
(x)(8-x)=-20 \\multiplying both sides by 1/4
x^2-8x-20=0 \\equating to zero gives
(x-10)(x+2)=0 \\the two factors of the equation

therefore:

If x=10 and k=-6, your sequence will be;
{10, 4, -2}
If x=-2 and k=6, your sequence will be;
{-2, 4, 10}

That's the answer. :)