# 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?

May 1, 2016

$23 , 27 , 31$

#### Explanation:

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:

${b}_{2}^{2} = {b}_{1} {b}_{3}$

${a}^{2} + 14 a + 49 = {\left(a + 7\right)}^{2} = \left(a + 2\right) \left(a + 13\right) = {a}^{2} + 15 a + 26$

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

$a = 23$

So ${a}_{1} , {a}_{2} , {a}_{3} = 23 , 27 , 31$