Question

Three numbers are in an arithmetic progression with a common difference of 6. if 4 is subtracted from the first number, 1 is subtracted from the second number, and the third number is first decreased by 3 and then multiplied by 3, then the resulting three numbers form a geometric progression. find the original three numbers.

Answer 1

Answer:

  either of {-9.5, -3.5, 2.5} or {7, 13, 19}

Step-by-step explanation:

Let x represent the first number of the arithmetic sequence. Then the second number is x+6, and the third number is x+12. The geometric sequence is then ...

  (x -4), (x +5), 3(x+9)

The common ratio is then ...

  (x +5)/(x -4) = 3(x +9)/(x +5)

Cross-multiplying gives ...

  (x +5)² = 3(x -4)(x +9)

  x² +10x +25 = 3x² +15x -108 . . . . eliminate parentheses

  0 = 2x² +5x -133 . . . . . . . . . . . . . . subtract the left side

  0 = (2x +19)(x -7) . . . . . . . . . . . . . . factor

  x = -9.5 . . or . . 7

The possible original three numbers are ...

  {-9.5, -3.5, 2.5} . .  or . . {7, 13, 19}