SOLUTION: Question: The sum of three numbers in a GP is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting sumber are in ARITHMETIC sequence.

Question 1163863: Question: The sum of three numbers in a GP is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting sumber are in ARITHMETIC sequence. Find the Geometric sequence.
How far I got:
a + ar + ar^2 = 14 (1)
(a+1), (ar+1), (ar^2 - 1) = AP with sum of 15 (2)
Since the(2) is an AP:(ar+1) = [(a+1) + (ar^2 - 1)]/ 2
(ar+1) = [a(1+ar^2)]/2
2(ar+1) = a(1+ar^2)
Am I on the right path or does it need some other approach?
Than you
Answer by ikleyn(52792) (Show Source): You can put this solution on YOUR website!
.

Your first equation is

    a + ar + ar^2 = 14     (1)


Let's will make the second equation.


Since the(2) is an AP :  (ar+1) = [(a+1) + (ar^2 - 1)]/ 2


Simplify it by canceling  "1" and "-1" inside the  [ . . . ].

    (ar+1) = [a + ar^2]/ 2

     2ar + 2 = a + ar^2

     a - 2ar + ar^2 = 2      (2)



Write equations (1) and (2) together

    a + ar + ar^2 = 14      (1)

    a - 2ar + ar^2 = 2      (2)


Subtract eq(2) from eq(1)

       3ar         = 12

        ar         = 12/3 = 4.    (3)


Now, in equation (1),  replace ar by 4  TWO TIMES,  based on (3).  You will get then

     a + 4 + 4r = 14     

     a + r      = 14-4 = 10.


Now you have two equations

    a + r = 10   
and
    ar    = 4.


You can solve it via substitution  a = 10 - r

    (10-r)*r = 4

    10r - r^2 = 4

    r^2 - 10r + 4 = 0


and so on . . .

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