Question

Three numbers form an arithmetic sequence whose
common difference is 3. If the first number is
increased by 1, the second increased by 6, and
the third increased by 19, the resulting three
numbers form a geometric sequence. Determine
the original three numbers.

Answer 1

Let x be the first number in the sequence. Then the first three numbers are

{x, x + 3, x + 6}

The next sentence says that the sequence

{x + 1, x + 9, x + 25}

is geometric, which means there is some fixed number r for which

x + 9 = r (x + 1)

x + 25 = r (x + 9)

Solve for r :

r = (x + 9)/(x + 1) = (x + 25)/(x + 9)

Solve for x :

(x + 9)² = (x + 25) (x + 1)

x ² + 18x + 81 = x ² + 26x + 25

8x = 56

x = 7

Then the three numbers are

{7, 10, 13}