Question 955519
<pre>
If the first term is "a" and the difference is "d", then the first four terms 
are 

a, a+d, a+2d, a+3d 
</pre>
the sum of first four terms of an arithmetic sequence is 16 
<pre>
So, 

a + a+d + a+2d + a+3d = 16
              4a + 6d = 16 

Divide thru by 2 

              2a + 3d = 8 
</pre>
their product 105, 
<pre>
So  
   a(a+d)(a+2d)(a+3d) = 105 

So we have this system of two equations: 

{{{system(2a + 3d = 8,a(a+d)(a+2d)(a+3d) = 105)}}} 

Solve the first equation for "a":

2a = 8-3d

 a = {{{(8-3d)/2}}}

Substitute in the second equation:

{{{a(a+d)(a+2d)(a+3d) = 105}}}

{{{((8-3d)/2)((8-3d)/2+d)((8-3d)/2+2d)((8-3d)/2+3d) = 105}}}

Get LCD in each parentheses:

{{{((8-3d)/2)((8-3d)/2+2d/2)((8-3d)/2+4d/2)((8-3d)/2+6d/2) = 105}}}

Combine fractions in parentheses:

{{{((8-3d)/2)((8-3d+2d)/2)((8-3d+4d)/2)((8-3d+6d)/2) = 105}}}

{{{((8-3d)/2)((8-d)/2)((8+d)/2)((8+3d)/2) = 105}}}

Multiply both sides by 16

{{{(8-3d)(8-d)(8+d)(8+3d) = 1680}}}

Rearrange the factors on the left:

{{{(8-3d)(8+3d)(8-d)(8+d) = 1680}}}

FOIL

{{{(64-9d^2)(64-d^2) = 1680}}}

FOIL again:

{{{4096-640d^2+9d^4=1680}}}

Get 0 on the right side:

{{{2416-640d^2-9d^4=0}}}

Get in descending order:

{{{9d^4-640d^2+2416=0}}}

Factor the trinomial:

{{{(9d^2-604)(d^2-4)=0}}}

Factor further:

{{{(9d^2-604)(d-2)(d+2)=0}}}

{{{9d^2-604=0}}}, {{{d-2=0}}}, {{{d+2=0}}}
{{{9d^2=604}}}, {{{d=2}}}, {{{d=-2}}}
{{{d^2=604/9}}}
{{{d= "" +- sqrt(604/9)}}}
{{{d= "" +- sqrt(4*151)/3}}}
{{{d= "" +- 2sqrt(151)/3}}}

We can ignore the irrational values for d, since the sequence 
has only integers.

The only integer values for d are d=2 and d=-2

For d=2, substitute in

 a = {{{(8-3d)/2}}}
 a = {{{(8-3*2)/2}}}
 a = {{{(8-6)/2}}}
 a = {{{2/2}}}
 a = {{{1}}}

Substitute d=2 and a=1 in

a, a+d, a+2d, a+3d

1, 1+2, 1+2(2), 1+3(2)

Therefore 1 solution is

1, 3, 5, 7

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

For d=-2, substitute in

 a = {{{(8-3d)/2}}}
 a = {{{(8-3*(-2))/2}}}
 a = {{{(8+6)/2}}}
 a = {{{14/2}}}
 a = {{{7}}}

Substitute d=-2 and a=7 in

a, a+d, a+2d, a+3d

7, 7+(-2), 7+2(-2), 7+3(-2)

7, 5, 7-4, 7-6

Therefore another solution is

7, 5, 3, 1.

But this is the same sequence reversed.

Edwin</pre>