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?

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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?

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Answer 1 · 2016-05-01T19:18:13

$23, 27, 31$ $23, 27, 31$

Explanation:

The original sequence is:

$a_{1} = a$ $a_1 = a$

$a_{2} = a + 4$ $a_2 = a+4$

$a_{3} = a + 8$ $a_3 = a+8$

The modified sequence is:

$b_{1} = a + 2$ $b_1 = a+2$

$b_{2} = a + 7$ $b_2 = a+7$

$b_{3} = a + 13$ $b_3 = a+13$

Since $b_{1}, b_{2}, b_{3}$ $b_1, b_2, b_3$ is a geometric sequence, the middle term $b_{2}$ $b_2$ must be a geometric mean of $b_{1}$ $b_1$ and $b_{3}$ $b_3$ . Hence:

$b_{2}^{2} = b_{1} b_{3}$ $b_2^2 = b_1 b_3$

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

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

$a = 23$ $a=23$

So $a_{1}, a_{2}, a_{3} = 23, 27, 31$ $a_1, a_2, a_3 = 23, 27, 31$

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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?

1 Answer

Explanation:

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