Question 1102616
<pre>
First term = a-d
Second term = a
Third term = a+d

{{{system((a-d)+a=30,(a-d)(a+d)=153)}}}

{{{system(a-d+a=30,a^2-d^2=153)}}}

{{{system(2a-d=30,a^2-d^2=153)}}}

Solve the first equation for d:

{{{2a-30=d}}}

Substitute 2a-30 for d in

{{{a^2-d^2=153}}}

{{{a^2-(2a-30)^2=153}}}

{{{a^2-(4a^2-120a+900)=153}}}

{{{a^2-4a^2+120a-900=153}}}

{{{-3a^2+120a-900=153}}}

Divide through by -3

{{{a^2-40a+300=-51}}}

Get 0 on the right:

{{{a^2-40a+351=0}}}

Factor:

{{{(a-13)(a-27)=0}}}

a-13=0 ;  a-27=0
   a=13;     a=27

Substitute each in

{{{2a-30=d}}}

{{{2(13)-30=d}}};   {{{2(27)-30=d}}}
{{{26-30=d}}};   {{{54-30=d}}}
{{{-4=d}}};   {{{24=d}}}

So one solution is

First term = a-d = 13-(-4) = 13+4 = 17
Second term = a = 13
Third term = a+d = 13+(-4) = 13-4 = 9

17,13,9

And the other solution is

First term = a-d = 27-(24) = 3
Second term = a = 27
Third term = a+d = 27+(24) = 51

3,27,51

In both cases, the sum of the first and second 
terms is 30, and the product of the first and 
third terms is 153.

Edwin</pre>