Question 840912
The first two terms of a geometric sequence and an arithmetic sequence are the same. The first term is 12. The sum of the first three terms of the geometric sequence is 3 more than the sum of the first three terms of the arithmetic sequence. Determine TWO possible values for the common ratio, r, of the geometric sequence
<pre>
Geometric sequence = a,ar,arē,__,__,...

Arithmetic sequence= a,a+d,a+2d,__,__,...</pre>The first two terms of a geometric sequence and an arithmetic sequence are the same.<pre>a = a, ar = a+d

The first term is 12. 

a = 12,  12r = 12+d</pre>The sum of the first three terms of the geometric sequence is 3 more than the sum of the first three terms of the arithmetic sequence.<pre>   (a)+(ar)+(arē) = (a)+(a+d)+(a+2d)+3
(12)+(12r)+(12rē) = (12)+(12+d)+(12+2d)+3
      12+12r+12rē = 12+12+d+12+2d+3
      12+12r+12rē = 39+3d
         12r+12rē = 27+3d

Every term can be divided by 3

           4r+4rē = 9+d

So we have this system of two equations and 2 unknowns:

     12r = 12+d, 4r+4rē = 9+d

Solve the first for d:

         12r = 12+d     
      12r-12 = d

Substitute in

      4r+4rē = 9+d
      4r+4rē = 9+(12r-12)
      4r+4rē = 9+12r-12
      4r+4rē = 12r-3
    4rē-8r+3 = 0
(2r-1)(2r-3) = 0
   2r-1=0;  2r-3=0
     2r=1;    2r=3
      r={{{1/2}}};    r={{{3/2}}}

Those are the two values.

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

To check we must find d

Using r = {{{1/2}}}

 12r-12 = d
12{{{(1/2)}}}-12 = d
   6-12 = d
     -6 = d

Arithmetic sequence:  12, 6, 0, ...
Geometric sequence:   12, 6, 3, ...

Sum of 1st three terms of arithmetic sequence = 12+6+0 = 18
Sum of 1st three terms of geometric sequence  = 12+6+3 = 21
21 is 3 more than 18
That checks.

Using r = {{{3/2}}}

 12r-12 = d
12{{{(3/2)}}}-12 = d
  18-12 = d
      6 = d

Arithmetic sequence:  12, 18, 24, ...
Geometric sequence:   12, 18, 27, ...

Sum of 1st three terms of arithmetic sequence = 12+18+24 = 54
Sum of 1st three terms of geometric sequence  = 12+18+27 = 57
57 is 3 more than 54
That checks, too.  So it's correct

Edwin</pre>