Question 224627
The first three terms of an arithmetic series have a sum of 24 and a product of 512. What is the fourth term in the series?
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Let: 
The common difference = d
The first term = a

Therefore:

The second term = a+d
The third term = a+d+d = a+2d

>>...The first three terms of an arithmetic series have a sum of 24...<<

           a + (a+d) + (a+2d) = 24
                   a+a+d+a+2d = 24
                        3a+3d = 24
                          a+d = 8
 
>>...The first three terms of an arithmetic series have...a product of 512...<<

                 a(a+d)(a+2d) = 512  
                 
So we have to solve the system:

{{{system(a+d=8,a(a+d)(a+2d)=512)}}}

Solve the first equation for d

{{{d=8-a}}}

Substitute in

{{{a(a+d)(a+2d) = 312}}}
{{{a(a+(8-a))(a+2(8-a)) = 312}}}
{{{a(a+8-a)(a+16-2a)) = 312}}}
{{{a(8)(-a+16) = 312}}}
{{{8a(-a+16) = 312}}}
Divide both sides by 8
{{{a(-a+16)=39
{{{-a^2+16a=39}}}
{{{-a^2+16a-39=0}}}
Multiply through by -1
{{{a^2-16a+39=0}}}
Factor:
{{{(a-13)(a-3)=0}}}
Set each factor = 0

{{{a-13=0}}},     {{{a-3=0}}}
{{{a=13}}}      {{{a=3}}}

There are two solutions for the first term, a,

Now since {{{d=8-a}}}, taking the first solution:

When {{{a=13}}}, {{{d=8-a=8-(13)=8-13=-5}}}

so the first three terms are

{{{13}}}, {{{13+(-5)=8}}}, {{{8+(-5)=3}}}

Checking:  their sum = 13+8+3=24
           their product = (13)(8)(3)=312

So the fourth term is the third term plus (-5), or

fourth term = 3-5 = -2

-----------------------------

As before, since {{{d=8-a}}}, taking the second solution:

When {{{a=3}}}, {{{d=8-a=8-(13)=8-3=5}}}

so the first three terms are

{{{3}}}, {{{3+5=8}}}, {{{8+5=13}}}

Checking:  their sum = 3+8+13=24
           their product = (3)(8)(13)=312

So the fourth term is the third term plus 5, or

fourth term = 13+5 = 18

So there are two solutions for the fourth term: -2, and 18

Edwin</pre>