Three numbers form an arithmetic sequence having a common difference of 4. If the first number is increased by 2, the second number by 3, and the 3rd number by 5, the resulting numbers form a geometric sequence. How do you find the original numbers?

Question

12371 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-01T19:18:13

$23, 27, 31$

The original sequence is:

$a_1 = a$

$a_2 = a+4$

$a_3 = a+8$

The modified sequence is:

$b_1 = a+2$

$b_2 = a+7$

$b_3 = a+13$

Since $b_1, b_2, b_3$ is a geometric sequence, the middle term $b_2$ must be a geometric mean of $b_1$ and $b_3$ . Hence:

$b_2^2 = b_1 b_3$

$a^2+14a+49 = (a+7)^2 = (a+2)(a+13) = a^2+15a+26$

Subtracting $a^2+14a+26$ from both ends, we find:

$a=23$

So $a_1, a_2, a_3 = 23, 27, 31$