80.7k views
1 vote
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.

User Overblade
by
8.2k points

1 Answer

5 votes

Answer:

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

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}

User Maxflow
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories