SOLUTION: Question: In a set of four numbers, the first three are in a geometric sequence and the last three are in an arithmetic sequence with a common difference of 6. If the first number

n               1        2        3        4


GP              a        ar       ar^2     


AP                       ar       ar+6    ar+12



First equation

      "the first number is the same as the fourth" :  a = ar + 12



Second equation : 

      the terms under n= 3 in the table are equal :  ar^2 = ar + 6.



The system of equations is


    a = ar + 12       (1)

    ar^2 = ar + 6     (2)


To solve the system, write equations (1) and (2) in equivalent form


    a - ar = 12      (1')

   ar - ar^2 = -6    (2')



Transform equation (2') to the form  r*(a-ar) = -6  and replace (a-ar) in it by 12, based on (1').


You will get then

    12r = -6,   which implies  r =  =  = -0.5.


Having it, substitute  r = 0.5 into equation (1'). You will get then

    a - (-0.5)a = 12

    a + 0.5a    = 12

    1.5a        = 12

       a        = 12/1.5 = 8.


So, a = 8, r = -0.5,


Knowing it, you can calculate all the terms in both progressions.